Maxima And Minima

f(x) = 4x² – 4x + 4 on R

= 4x² – 4x + 1 + 3

= (2x – 1)² + 3

Since, (2x – 1)² ≥0

= (2x – 1)² + 3 ≥3

= f(x) ≥ f

Thus, the minimum value of f(x) is 3 at x =

Since, f(x) can be made large. Therefore maximum value does not exist.

We have f(x) = – (x – 1)² + 2

It can be observed that (x – 1)²≥0 for every x∈R

Therefore, f(x) = – (x – 1)² + 2≤2 for every x∈R

The maximum value of f is attained when (x – 1) = 0

(x – 1)=0, x=1

Since, Maximum value of f = f(1) = – (1 – 1)² + 2 = 2

Hence, function f does not have minimum value.

|x + 2|≥0 for x ∈ R

= f(x) ≥ 0 for all x ∈ R

So the minimum value of f(x) is 0, which attains at x =2

Hence, f(x) = |x + 2| does not have the maximum value.

We know that – 1 ≤ sin2x ≤ 1

= – 1 + 5 ≤ sin2x + 5 ≤ 1 + 5

= 4 ≤ sin 2x + 5 ≤ 6

Hence, the maximum and minimum value of h are 4 and 6 respectively.

We know that – 1 ≤ sin4x ≤ 1

= 2 ≤ sin4x + 3 ≤ 4

= 2 ≤ |sin 4x + 3| ≤ 4

Hence, the maximum and minimum value of f are 4 and 2 respectively.

We have f(x) = 2x³ + 5 on R

Here, we observe that the values of f(x) increase when the values of x are increased and f(x) can be made large,

So, f(x) does not have the maximum value

Similarly, f(x) can be made as small as we want by giving smaller values to x.

So, f(x) does not have the minimum value.

We know that – |x + 1| ≤ 0 for every x ∈ R.

Therefore, g(x) = – |x + 1| + 3 ≤ 3 for every x ∈ R.

The maximum value of g is attained when |x + 1| = 0

|x + 1| = 0

x = – 1

Since, Maximum Value of g = g( – 1) = – | – 1 + 1| + 3 = 3

Hence, function g does not have minimum value.

We have f(x) = 16x² – 16x + 28 on R

= 16x² – 16x + 4 + 24

= (4x – 2)² + 24

Now, (4x – 2)² ≥ 0 for all x ∈ R

= (4x – 2)² + 24≥ 24 for all x ∈ R

= f(x) ≥ f

Thus, the minimum value of f(x) is 24 at x =

Hence, f(x) can be made large as possibly by giving difference value to x.

Thus, maximum values does not exist.

We have f(x) = x³ – 1 on R

Here, we observe that the values of f(x) increase when the values of x are increased, and f(x) can be made large, by giving large value.

So, f(x) does not have the maximum value

Similarly, f(x) can be made as small as we want by giving smaller values to x.

So, f(x) does not have the minimum value.

f(x) = (x – 5)⁴

Differentiate w.r.t x

f ’(x) = 4(x – 5)³

for local maxima and minima

f ‘ (x) = 0

= 4(x – 5)³ = 0

= x – 5 = 0

x = 5

f ‘ (x) changes from –ve to + ve as passes through 5.

So, x = 5 is the point of local minima

Thus, local minima value is f(5) = 0

We have, g (x) = x³ – 3x

Differentiate w.r.t x then we get,

g’ (x) = 3x² – 3

Now, g‘(x) =0

= 3x² = 3 ⇒ x = ±1

Again differentiate g’(x) = 3x² – 3

g’’(x)= 6x

g’’(1)= 6 > 0

g’’( – 1)= – 6>0

By second derivative test, x=1 is a point of local minima and local minimum value of g at

x =1 is g(1) = 1³ – 3 = 1 – 3 = – 2

However, x = – 1 is a point of local maxima and local maxima value of g at

x = – 1 is g( – 1) = ( – 1)³ – 3( – 1)

= – 1 + 3

= 2

Hence, The value of Minima is – 2 and Maxima is 2.

We have, f(x) = x³(x – 1)²

Differentiate w.r.t x, we get,

f ‘(x) = 3x²(x – 1)² + 2x³(x – 1)

= (x – 1)(3x²(x – 1) + 2x³)

= (x – 1)(3x³ – 3x² + 2x³)

= (x – 1)(5x³ – 3x²)

= x² (x – 1)(5x – 3)

For all maxima and minima,

f ’(x) = 0

= x²(x – 1)(5x – 3) = 0

= x =0, 1,

At x = f ’(x) changes from –ve to + ve

Since, x = is a point of Minima

At x =1 f ‘ (x) changes from –ve to + ve

Since, x =1 is point of maxima.

We have, f(x) = (x – 1)(x + 2)²

Differentiate w.r.t x, we get,

f ‘(x) = (x + 2)² + 2(x – 1)(x + 2)

= (x + 2)(x + 2 + 2x – 2)

= (x + 2)(3x)

For all maxima and minima,

f ’(x) = 0

= (x + 2)(3x) = 0

= x =0, – 2

At x = – 2 f ’(x) changes from –ve to + ve

Since, x = – 2 is a point of Maxima

At x =0 f ‘ (x) changes from –ve to + ve

Since, x =0 is point of Minima.

Hence, local min value = f(0) = – 4

local max value = f( – 2) = 0.

We have, f(x) = (x – 1)³(x + 1)²

Differentiate w.r.t x, we get,

f ‘(x) = 3(x – 1)²(x + 1)² + 2(x – 1)³(x + 1)

= (x – 1)²(x + 1) (3 (x + 1) + 2(x – 1)

= (x – 1)²(x + 1)(5x + 1)

For the point of local maxima and minima,

f ’(x) = 0

= (x – 1)²(x + 1)(5x + 1) = 0

= x = 1, – 1,

At x = – 1 f ’(x) changes from –ve to + ve

Since, x = – 1 is a point of Maxima

At x = f ‘ (x) changes from –ve to + ve

Since, x = is point of Minima.

Hence, local min value =

local max value = 0.

We have, f(x) = x³ – 6x² + 9x + 15

Differentiate w.r.t x, we get,

f ‘(x) = 3x² – 12x + 9

= 3(x² – 4x + 3)

= 3(x – 3)(x – 1)

For all maxima and minima,

f ’(x) = 0

= 3(x – 3)(x – 1) = 0

= x = 3, 1

At x = 1 f ’(x) changes from –ve to + ve

Since, x = – 1 is a point of Maxima

At x =3 f ‘ (x) changes from –ve to + ve

Since, x =3 is point of Minima.

Hence, local max value f(1)= (1)³ – 6(1)² + 9(1) + 15 = 19

local min value f(3) = (3)³ – 6(3)² + 9(3) + 15 = 15

We have, f(x) = sin 2x

Differentiate w.r.t x, we get,

f ‘(x) = 2cos 2x, 0 < x,π

For, the point of local maxima and minima,

f ’(x) = 0

= 2x =

= x =

At x = f ’(x) changes from –ve to + ve

Since, x = is a point of Maxima

At x = f ‘ (x) changes from –ve to + ve

Since, x = is point of Minima.

Hence, local max value f = 1

local min value f = – 1

We have, f(x) = sin x – cos x

Differentiate w.r.t x, we get,

f ‘(x) = cos x + sin x

For, the point of local maxima and minima,

f ’(x) = 0

= cos x = – sin x => tan x = – 1 = x =

Again differentiate w.r.t x

f ’’(x) = – sin x + cos x

f ”= – sin + cos = –

f ”= – sin + cos = –

Therefore, by second derivative test, x = is a point of local maxima and the local maximum of f at x = is

F” = sin – cos =

F” = sin – cos =

f(x) = cos x

Differentiate w.r.t x

f ’(x) = – sin x

for the point of local maxima and minima,

f ‘ (x) = 0

= – sin x = 0

x = 0, and π

But, these two interval lies outside the interval (0,π)

So, no local maxima and minima will exist in the interval (0, π)

We have, f(x) = sin 2x – x

Differentiate w.r.t x, we get,

f ‘(x) = 2cos 2x – 1,

For, the point of local maxima and minima,

f ’(x) = 0

2cos2x – 1 = 0

= cos 2x = = cos

= 2x =

= x =

At x = f ’(x) changes from –ve to + ve

Since, x = is a point of Maxima

At x = f ‘ (x) changes from –ve to + ve

Since, x = is point of local maxima

Hence, local max value f

local min value f

We have, f(x) = 2 sin x – x

Differentiate w.r.t x, we get,

f ‘(x) = 2cos x – 1 = 0

For, the point of local maxima and minima,

f ’(x) = 0

cos x =

= x =

At x = f ’(x) changes from –ve to + ve

Since, x = is a point of Minima with value =

At x = f ‘ (x) changes from –ve to + ve

Since, x = is point of local maxima with value =

Hence, local max value f

local min value f

=

=

For f ‘(0) =

So, 2 – 3x = 0 => x =

=

=

=

Therefore, by second derivative test, x = is a point of local maxima and the local maxima value of F at x = is

Hence,

We have, f(x) = x³(2x – 1)³

Differentiate w.r.t x, we get,

f ‘(x) = 3x²(2x – 1)³ + 3x³(2x – 1)^2.2

= 3x²(2x – 1)²(2x – 1 + 2x)

= 3x²(4x – 1)

For the point of local maxima and minima,

f ’(x) = 0

= 3x²(4x – 1)= 0

= x = 0,

At x = f ’(x) changes from –ve to + ve

Since, x = is a point of Minima

Hence, local min value f =

We have, f(x) =

Differentiate w.r.t x, we get,

f ‘(x) =

For the point of local maxima and minima,

f ’(x) = 0

=

= x² – 4 = 0

= x =

= x = 2, – 2

At x = 2 f ’(x) changes from –ve to + ve

Since, x = 2 is a point of Minima

Hence, local min value f (2) = 2

f(x) = x⁴ – 62x^{2 +} 120x + 9

∴ f'(x) = 4x³ – 124x + 120 = 4(x³ – 31x + 30)

f''(x) = 12x² – 124 = 4(3x² – 31)

for maxima and minima,

f'(x) = 0

4(x³ – 31x + 30) = 0

So roots will be x = 5, 1, – 6

Now,

f''(5) = 176 > 0

x = 5 is point of local minima

f''(1) = – 112 < 0

x = 1 is point of local maxima

f''(– 6) = 308 > 0

x = – 6 is point of local minima

local max value = f(1) = 68

local min value = f(5) = – 316

and f(– 6) = – 1647

f(x) = x³ – 6x² + 9x + 15

∴ f'(x) = 3x² – 12x + 9 = 3(x² – 4x + 3)

f''(x) = 6x – 12 = 6(x – 2)

for maxima and minima,

f'(x) = 0

3(x² – 4x + 3) = 0

So roots will be x = 3, 1

Now,

f''(3) = 6 > 0

x = 3 is point of local minima

f''(1) = – 6 < 0

x = 1 is point of local maxima

local max value = f(1) = 19

local min value = f(3) = 15

f(x) = (x – 1) (x + 2)²

∴ f'(x) = (x + 2)² + 2(x – 1)(x + 2)

= (x + 2)(x + 2 + 2x – 2)

= (x + 2)(3x)

And f''(x) = 3(x + 2) + 3x

= 6x + 6

for maxima and minima,

f'(x) = 0

(x + 2)(3x) = 0

So roots will be x = 0, – 2

Now,

f''(0) = 6 > 0

x = 0 is point of local minima

f''(– 2) = – 6 < 0

x = 1 is point of local maxima

local max value = f(– 2) = 0

local min value = f(0) = – 4

we have f(x) = , x > 0

And,

for maxima and minima,

f'(x) = 0

x = 2

now,

f''(2) =

x = 2 is point of local maxima

local max value = f(2) = 1/2

we have f(x) = x e^x

f'(x) = e^x + xe^x = e^x(x + 1)

f''(x) = e^x(x + 1) + e^x

= e^x(x + 2)

For maxima and minima,

f'(x) = 0

e^x(x + 1) = 0

x = – 1

now f''(– 1) = e ^{– 1} = 1/e > 0

x = – 1 is point of local minima

hence, local min = f(– 1) = – 1/e

f(x) = , x > 0

f'(x) =

And f’’(x) =

For maxima and minima,

f'(x) = 0

= 0

x = 2, – 2

now,

f’’(2) = 1/2 > 0

x = 2 is point of minima

we will not consider x = – 2 as x > 0

local min value = f(2) = 2

we have f(x) = (x + 1) (x + 2)^1/3, x ≥ –2

f’(x) = (x + 2)^1/3 + 1/3(x + 1)(x + 2) ^{– 2/3}

= (x + 2) ^{– 2/3}(x + 2 + 1/3(x + 1))

= 1/3(x + 2) ^{– 2/3}(4x + 7)

And f’’(x) =

For maxima and minima,

f'(x) = 0

1/3(x + 2) ^{– 2/3}(4x + 7) = 0

x = – 7/4

Now

f’’(– 7/4) =

x = – 7/4 is point of minima

local min value = f(– 7/4) =

we have,

And f’’(x) =

For maxima and minima,

f'(x) = 0

X = ±4

Now

f’’(4) =

x = 4 is point of maxima

local maximum value = f(4)

= 4√(32 – 4²)

= 4√(32 – 16)

= 4√16

= 16

Local minimum at x = – 4

Local minimum value = f(– 4)

= – 4√(32 – (– 4)²)

= – 4√(32 – 16)

= – 4√16

= – 16

: f(x) = , a > 0 x≠0

f'(x) =

and f’’(x) =

For maxima and minima,

f'(x) = 0

= 0

x = ±a

now,

f’’(a) = 2/a > 0 as a > 0

x = a is point of minima

f’’(– a) = – 2/a < 0 as a > 0

x = – a is point of maxima

hence

local max value = f(– a) = – 2a

local min value = f(a) = 2a

f(x) = x

f’(x) =

f’’(x) =

For maxima and minima,

f'(x) = 0

X = ±1

Now

f’’(1) < 0

x = 1 is point of local maxima

f’’(– 1) > 0

x = – 1 is point of minima

hence

local max value = f(1) = 1

local min value = f(– 1) = – 1

f’(x) =

f’’(x) =

For maxima and minima,

f'(x) = 0

X = 1 – 1/4 = 3/4

Now

f’’(3/4) < 0

x = 3/4 is point of local maxima

hence

local max value = f(3/4) = 5/4

f(x) = (x – 1)(x – 2)2

f’(x) = (x – 2)² + 2(x – 1)(x – 2)

= (x – 2)(x – 2 + 2x – 2)

= (x – 2)(3x – 4)

f’’(x) = (3x – 4) + 3(x – 2)

For maxima and minima,

f'(x) = 0

(x – 2)(3x – 4) = 0

x = 2, 4/3

Now

f’’(2) > 0

x = 2 is point of local minima

f’’(4/3) = – 2 < 0

x = 4/3 is point of local maxima

hence

local max value = f(4/3) = 4/27

local min value = f(2) = 0

f’’(x) =

For maxima and minima,

f'(x) = 0

x = 2/3

Now

f’’(2/3) < 0

x = 2/3 is point of maxima

hence

local max value = f(2/3) =

f(x) = – (x – 1)³(x + 1)²

f’(x) = – 3(x – 1)²(x + 1)² – 2(x – 1)³(x + 1)

= – (x – 1)²(x + 1)(3x + 3 + 2x – 2)

= – (x – 1)²(x + 1)(5x + 1)

f’’(x) = – 2(x – 1)(x + 1)(5x + 1) – (x – 1)²(5x + 1) – 5(x – 1)²(x – 1)

For maxima and minima,

f'(x) = 0

– (x – 1)²(x + 1)(5x + 1) = 0

x = 1, – 1, – 1/5

Now

f’’(1) = 0

x = 1 is inflection point

f’’(– 1) = – 4× – 4 = 16 > 0

x = – 1 is point of minima

f’’(– 1/5) = – 5(36/25)*4/5 = – 144/25 < 0

x = – 1/5 is point of maxima

hence

local max value = f(– 1/5) =

local min value = f(– 1) = 0

we have

y = a log x + bx² + x

And

For maxima and minima,

Given that extreme value exist at x = 1, 2

a + 2b = – 1 ……(1)

a + 8b = – 2 …… (2)

solving (1) and (2) we get

a = – 2/3 b = – 1/6

the given function is

Now f’(x) = 0

1 – log x = 0

log x = 1

log x = log e

x = e

now

Now f’’(x) =

Therefore, by second derivation test f is the maximum at x = e.

f’(x) =

f’’(x) =

For maxima and minima,

f'(x) = 0

(x + 2)² = 4

x² + 4x = 0

x(x + 4) = 0

x = 0, – 4

Now f’’(0) = 1 > 0

x = 0 is point of minima

f’’(– 4) = – 1 < 0

x = – 4 is point of maxima

Local max value = f(– 4) = – 6

Local min value = f(0) = 2

we have

y = tanx – 2x

y’ = sec²x – 2

y’’ = 2sec²xtanx

For maxima and minima,

y’ = 0

sec2x = 2

secx = ±√2

x =

y’’() = 4 > 0

x = is the point of minima

y’’() = – 4 < 0

x = is the point of maxima

hence

max value = f() = – 1 –

min value = f() = 1 –

consider the function f(x) = x³ + ax² + bx + c

Then f’(x) = 3x² + 2ax + b

It is given that f(x) is maximum at x = – 1

f’(– 1) = 3(– 1)2 + 2a(– 1) + b = 0

f’(– 1) = 3 – 3a + b = 0 ……(1)

it is given that f(x) is minimum at x = 3

f’(x) = 3(3)2 + 2a(3) + b = 0

f’(3) = 27 + 6a + b = 0 …… (2)

solving equation (1) and (2) we have

a = – 3 and b = – 9

since f’(x) is independent of constant c, it can be any real number

f(x) = sin x + √3 cos x

f’(x) = cos x – √3 sin x

Now,

f’(x) = 0

cos x – √3 sin x = 0

cos x = √3 sin x

cot x = √3

x =

Differentiate f’’(x), we get

f’’(x) = – sin x –√3 cos x

f’’() =

Hence, at x = is the point of local maxima.

given function is f(x) =

∴f'(x) = 4 – x

Now,

f'(x) = 0

4 – x = 0

x = 4

Then, we evaluate of f at critical points x = 4 and at the interval [ – 2, ]

f(4) = = 8

f(– 2) =

f() =

Hence, we can conclude that the absolute maximum value of f on [ – 2, 9/2] is 8 occurring at x = 4 and the absolute minimum value of f on [ – 2, 9/2] is – 10 occurring at x = – 2

given function is f(x) =

∴f'(x) = 2(x – 1)

Now,

f'(x) = 0

2(x – 1) = 0

x = 1

Then, we evaluate of f at critical points x = 1 and at the interval [ – 3, 1]

f(1) = (1 – 1)² + 3 = 3

f(– 3) = (– 3 – 1)² + 3 = 19

Hence, we can conclude that the absolute maximum value of f on [ – 3, 1] is 19 occurring at x = – 3 and the minimum value of f on [ – 3, 1] is 3 occurring at x = 1

given function is f(x) =

Now,

f'(x) = 0

x = 2 or x² + 2 = 0 for which there are no real roots.

Therefore, we consider only x = 2 [0, 3].

Then, we evaluate of f at critical points x = 2 and at the interval [0, 3]

f(2) =

f(2) = 48 – 64 + 48 – 96 + 25 = – 39

f(0) =

f(3) =

Hence, we can conclude that the absolute maximum value of f on [0, 3] is 25 occurring at x = 0 and the minimum value of f on [0, 3] is – 39 occurring at x = 2

f(x) = (x – 2)

f’(x) =

put

f’(x) = 0

⇒

⇒

⇒

⇒

Now,

f(1) = 0

f(4/3) =

Hence, we can conclude that the absolute maximum value of f is 14 occurring at x = 9 and the minimum value of is occurring at x = 4/3

let f(x) = 2x³ – 24 x + 107

∴ f'(x) = 6 x² – 24 = 6(x² – 4)

Now,

f'(x) = 0

⇒ 6(x² – 4) = 0

⇒ x² = 4

⇒ x = ±2

We first consider the interval [1, 3].

Then, we evaluate the value of f at the critical point x = 2 [1, 3] and at the end points of the interval [1, 3].

f(2) = 2 (2³)– 24 (2) + 107 = 75

f(1) = 2(1)³ – 24(1) + 107 = 85

f(3) = 2(3)³– 24 (3) + 107 = 89

Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3,

Next, we consider te interval [ – 3, – 1].

Evaluate the value of f at the critical point x = – 2 [1, 3]

f(x) = cos²x + sin x

f’(x) = 2 cos x (–sin x) + cos x

= 2 sin x cos x + cos x

now, f’(x) = 0

⇒ 2 sin x cos x = cos x

⇒ cos x(2sin x – 1) = 0

⇒ sin x = 1/2 or cos x = 0

⇒ x = or as x [0, ]

So, the critical points are x = and x = and at the end point of the interval [0, ] we have,

f() = = 5/4

f(0) = = 1 + 0 = 1

f(π) = = (–1²) + 0 = 1

f() = = 0 + 1 = 1

Thus, we conclude that the absolute maximum value of f is 5/4 at x = , and absolute minimum value of f is 1 which occurs at x = 0, .

We have,

f(x) =

f’(x) =

Thus, f’(x) = 0

⇒ x =

Further note that f’(x) is not define at x = 0.

So, the critical points are x = 0 and x = and at the end point of the interval x = – 1 and x = 1

f(– 1) =

f(0) =

f() =

f(1) =

Thus, we conclude that the absolute maximum value of f is 18 at x = 1, and absolute minimum value of f is which occurs at x = .

Given,

f (x) = 2x³ – 15x² + 36x + 1

f’(x) = 6 x² – 30 x + 36

f’(x) = 6(x² – 5 x + 6) = 6 (x – 2)(x – 3)

Note that f’(x) = 0 gives x = 2 and x = 3

We shall now evaluate the value of f at these points and at the end points of the interval [1, 5],

i.e. at x = 1, 2, 3 and 5

At x = 1, f(1) = 2(1³) – 15(1)² + 36(1) + 1 = 24

At x = 2, f(2) = 2(2³) – 15(2)² + 36(2) + 1 = 29

At x = 3, f(3) = 2(3)³– 15(3)²+ 36(3) + 1 = 28

At x = 5, f(5) = 2(5)³– 15(5)²+ 36(5) + 1 = 56

Thus, we conclude that the absolute maximum value of f on [1, 5] is 56, occurring at x = 5, and absolute value of f on [1, 5] is 24 which occurs at x = 1.

Let the two positive numbers be a and b.

Given: a + b = 15 … 1

Also, a² + b² is minima

Assume, S = a² + b²

(from equation 1)

⇒ S = a² + (15 – a)²

⇒ S = a² + 225 + a² – 30a = 2a² – 30a + 225

⇒ = 4a - 30

⇒

Since, > 0 will give minimum value of S.

4a – 30 = 0

a = 7.5

Hence, two numbers will be 7.5 and 7.5.

Let the two positive numbers be a and b.

Given: a + b = 64 … 1

Also, a³ + b³ is minima

Assume, S = a³ + b³

(from equation 1)

S = a³ + (64 – a)³

= 3a² + 3(64 – a)² × ( - 1)

(condition for maxima and minima)

⇒ 3a² + 3(64 – a)² × ( - 1) = 0

⇒ 3a² + 3(4096 + a² – 128a) × ( - 1) = 0

⇒ 3a² – 3 × 4096 - 3a² + 424a = 0

⇒ a = 32

Since, > 0 ⇒ a= 32 will give minimum value

Hence, two numbers will be 32 and 32.

Let a and b be two numbers such that a, b ≥ - 2

Given: a + b =

Assume, S = a + b³

S = a + ( - a)³

= 1 + 3( - a)²( - 1)

Condition maxima and minima is

For S to minimum,

Hence, and

Let the two numbers be a and b.

Given: a + b = 15

Assume, S = a²b³

S = a²(15 - a)³

= 2a(15 - a)³ - 3a²(15 - a)²

Condition maxima and minima is

⇒ 2a(15 - a)³ - 3a²(15 - a)² = 0

⇒ a(15 - a)² {2(15 – a) – 3a} = 0

⇒ a(15 - a)² {30 – 5a} = 0

⇒ a = 0, 15, 6

= 2(15 - a)³ - 6a(15 - a)² - 6a(15 - a)²+ 3a²(15 - a)

= (15 - a){2(15 - a)² - 12a(15 - a)+ 3a²}

= (15 - a){2(15 - a)² - 12a(15 - a)+ 3a²}

For S to minimum, > 0

a = 0 minima

a = 6 maxima

a = 15 point of inflection

Hence, two numbers are 0 and 15

Let the radius and height of right circular cylinder be r and h respectively.

Given: Volume of Cylindrical can = 100 cm³

Volume of a cylinder = πr²h

πr²h = 100

… 1

Surface of a cylinder, S = 2πrh + 2πr²

From equation we get,

Condition for maxima and minima

So, for

>0

This is the condition for minima

From equation 1,

Hence, required dimensions of cylinders are radius = and height =

Condition for maxima and minima is

And for M to maximum < 0.

(i)

<0

Hence, for , M will be maximum.

(ii)

= 0

So,

For

(condition for maximum value)

For

(condition for minimum value)

Therefore, for , M will have maximum value.

Let the length of side of square be a and radius of circle be r.

It is given that wire is cut into two parts to form a square and a circle

Therefore, perimeter of square + circumference of circle = length of wire

4a + 2πr = 28

a = …1

Let us assume area of square + area of circle = S

S = a² + πr²

S = + πr² (from equation 1)

Condition for maxima and minima

…2

So, for >0

This is the condition for minima

From equation 1

a =

Substituting from equation 2

a =

a =

a =

Hence, radius of circle and length of square be and respectively.

Let the length of side of square and equilateral triangle be a and r respectively.

It is given that wire is cut into two parts to form a square and a equilateral triangle

Therefore, perimeter of square + perimeter of equilateral triangle = length of wire

4a + 3r = 20

…1

Let us assume area of square + area of circle = S

S = a² +

S = + (from equation 1)

Condition for maxima and minima

= 0

…2

So, for >0

This is the condition for minima

From equation 1

Substituting from equation 2

Hence, side of equilateral triangle and length of square be and respectively.

Let us say the sum of perimeter of square and circumference of circle be L

Given: Sum of the perimeters of a square and a circle.

Assuming, side of square = a and radius of circle = r

Then, L = 4a + 2πr …1

Let the sum of area of square and circle be S

So, S = a² + πr²

S =

Condition for maxima and minima

= 0

So, for >0

This is the condition for minima

From equation 1

Substituting from equation 2

a = 2r

Hence, proved.

Given: Hypotenuse = 5 cm.

Let the two sides of right angle triangle be a and r.

According to the Pythagoras theorem, a² + r² = 25 …1

Assuming, area of triangle be,

Condition for maxima and minima

= 0

= 0

r² = 25 – r²

2r² = 25

Since, r cannot be negative

Therefore,

So, for ,

Area will be maximum

From equation 1,

a² + r² = 25

It is given that Two sides of a triangle have lengths ‘a’ and ‘b’ and the angle between them is θ.

Let the area of triangle be A

Then,

Condition for maxima and minima is

So, for A to be maximum,

For

Hence, will give maximum area.

And maximum area will be

Side length of big square is 18 cm

Let the side length of each small square be a.

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with

Length, L = 18 – 2a

Breadth, B = 18 – 2a and

Height, H = a

Assuming,

Volume of box, V = LBH = a(18 - 2a)²

Condition for maxima and minima is

(18 - 2a)² + (a)(- 2)(2)(18 - 2a) = 0

(18 - 2a)[(18 - 2a) - 4a] = 0

(18 - 2a)[18 – 6a] = 0

a = 3, 9

( - 2)(18 – 6a) + ( - 6)(18 – 2a)

For a = 3, = - 72,

For a = 9, = 72,

So for A to maximum

Hence, a = 3 will give maximum volume.

And maximum volume, V = a(18 - 2a)² = 432 cm³

Length of rectangle sheet = 45 cm

Breath of rectangle sheet = 24 cm

Let the side length of each small square be a.

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with

Length, L = 45 – 2a

Breadth, B = 24 – 2a and

Height, H = a

Assuming,

Volume of box, V = LBH = (45 - 2a)(24 - 2a)(a)

Condition for maxima and minima is

(45 - 2a)(24 - 2a) + ( - 2)(24 - 2a)(a) + (45 - 2a)( - 2)(a) = 0

4a² – 138a + 1080 + 4a² – 48a + 4a² – 90a = 0

12a² – 276a + 1080= 0

a² – 23a + 90= 0

a = 5, 18

= 24a - 276

For a = 5, = - 156,

For a = 18, = +156,

So for A to maximum

Hence, a = 5 will give maximum volume.

And maximum volume V = (45 - 2a)(24 - 2a)(a) = 2450 cm³

Let the length, breath and height of tank be l, b and h respectively.

Also, assume volume of tank as V

h = 2 m (given)

V = 8 cm³

lbh = 8

2lb = 8 (given)

lb = 4

b = …1

Cost for building base = Rs 70/m²

Cost for building sides = Rs 45/m²

Cost for building the tank, C = Cost for base + cost for sides

C = lb × 70 + 2(l + b) h × 45

C = l × × 70 + 2(l+) × 2 × 45

C = 280 + 180(l+) …2

Condition for maxima and minima

180(1 - ) = 0

l² = 4

l = ±2 cm

Since, l cannot be negative

So, l = 2 cm

For l = 2

Therefore, cost will be minimum for l =2

From equation 2

C = 280 + 180(l+)

C = Rs 1000

Let the radius of semicircle, length and breadth of rectangle be r, x and y respectively

AE = r

AB = x = 2r (semicircle is mounted over rectangle) …1

AD = y

Given: Perimeter of window = 10 m

x + 2y + πr = 10

2r + 2y + πr = 10

2y = 10 – (π + 2).r

y = … 2

To admit maximum amount of light, area of window should be maximum

Assuming area of window as A

A = xy +

A = (2r) () +

A = 10r - πr² – 2r² +

A = 10r – 2r² -

Condition for maxima and minima is

10 – 4r - πr = 0

= - 4 - π < 0

For r = A will be maximum.

Length of rectangular part = (from equation 1)

Breath of rectangular part = (from equation 2)

Let the side of equilateral triangle, length and breadth of rectangle be a, x and y respectively

AE = AB = a (ABE is equilateral triangle)

AB = x = a (triangle is mounted over rectangle) …1

AD = y

Perimeter of window = 12 m (given)

AE + EB + BC + CD + DA = 12

a + a + y + x + y = 10

2a + 2y + x = 10

3x + 2y = 12 (from equation 1)

… 2

To admit maximum amount of light, area of window should be maximum

Assuming area of window as A

A = xy +

(from equation 1 & 2)

Condition for maxima and minima is

= 0

x =

x =

x =

For x = A will be maximum.

Length of rectangular part = (from equation 1)

Breath of rectangular part = (from equation 2)

Let the radius, height and volume of cylinder be r, h and V respectively

Radius of sphere = R (Given)

Volume of cylinder, V = πr²h …1

OC = R

BC = r

In triangle OBC,

+ r² = R²

r² = R² - …2

Replacing equation 2 in equation 1, we get

V = π (R² - )(h) = πR²h -

Condition for maxima and minima is

= 0

Since, h cannot be negative

Hence,

For < 0

V will be maximum for

Let the length and breadth of rectangle ABCD be 2x and y respectively

Radius of semicircle = r (given)

In triangle OBA

r² = x² + y² (Pythagoras theorem)

y² = r² - x²

…1

Let us say, area of rectangle = A =xy

A = x () (from equation 1)

Condition for maxima and minima is

r² – x² = x²

2x² = r²

x =

Since, x cannot be negative

Hence,

For , < 0

A will be maximum for

From equation 1

y = =

Length of rectangle =

Breadth of rectangle =

Area of rectangle =

Let the radius and height of cone be r and h respectively

It is given that volume of cone is fixed.

Volume of cone, V =

h = …1

Curved surface area of cone, S = πrl (l is slant height)

Since,

So,

Condition for maxima and minima is

…2

For , > 0

S will be minimum for

From equation 1

(from equation 3)

h = √2 r

Let the radius and height of cone be r and h respectively

Radius of sphere = R

R² = r² + (h - R)²

R² = r² + h² + R² - 2hR

r² = 2hR - h² …1

Assuming volume of cone be V

Volume of cone, (from equation 1)

Condition for maxima and minima is

⇒ 4hR - 3 h²= 0

For , < 0

V will be maximum for

Let ‘r’ be the radius of the base circle of the cone and ‘l’ be the slant length and ‘h’ be the height of the cone:

Let us assume ‘’ be the semi - vertical angle of the cone.

We know that Volume of a right circular cone is given by:

⇒

Let us assume r²h = k(constant) …… (1)

⇒

⇒ …… (2)

We know that surface area of a cone is

⇒ …… (3)

From the cross - section of cone we see that,

⇒

⇒ …… (4)

Substituting (4) in (3), we get

⇒

From (2)

⇒

⇒

⇒

⇒

⇒

Let us consider S as a function of R and We find the value of ‘r’ for its extremum,

Differentiating S w.r.t r we get

⇒

Differentiating using U/V rule

⇒

⇒

⇒

⇒

⇒

⇒

Equating the differentiate to zero to get the relation between h and r.

⇒

⇒

Since the remainder is greater than zero only the remainder gets equal to zero

⇒ 2r⁶ = k²

From(1)

⇒ 2r⁶ = (r²h)²

⇒ 2r⁶ = r⁴h²

⇒ 2r² = h²

Since height and radius cannot be negative,

⇒ …… (5)

From the figure

⇒

From(5)

⇒

⇒

∴ Thus proved.

Δ ABC is an isosceles triangle such that AB = AC.

The vertical angle BAC = 2θ

Triangle is inscribed in the circle with center O and radius a.

Draw AM perpendicular to BC.

Since, Δ ABC is an isosceles triangle, the circumcenter of the circle will lie on the perpendicular from A to BC.

Let O be the circumcenter.

BOC = 2×2θ = 4θ (Using central angle theorem)

COM = 2θ (Since, Δ OMB and Δ OMC are congruent triangles)

OA = OB = OC = a (radius of the cicle)

In Δ OMC,

CM = asin2θ

OM = acos2θ

BC = 2CM (Perpendicular from the center bisects the chord)

BC = 2asin2θ

Height of Δ ABC = AM = AO + OM

AM = a + acos2θ

Differentiation this equation with respect to θ

Maxima or minima exists when:

Therefore,

Therefore,

To check whether which point has a maxima, we have to check the double differentiate.

Therefore, at θ = :

Both the sin values are positive. So the entire expression is negative. Hence there is a maxima at this point.

θ = will not form a triangle. Hence it is discarded.

There fore the maxima exits at:

Let PQR is the triangle with inscribed circle of radius ‘r’, touching sides PQ at Y, QR at X and PR at Z.

OZ, OX, OY are perpendicular to the sides PR,QR,PQ.

Here PQR is an isosceles triangle with sides PQ = PR and also from the figure,

⇒ PY = PZ = x

⇒ YQ = QX = XR = RZ = y

From the figure we can see that,

⇒ Area(ΔPQR) = Area(ΔPOR) + Area(ΔPOQ) + Area(ΔQOR)

We know that area of a triangle = ×base×height

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ …… (1)

We know that perimeter of the triangle is Per = PQ + QR + RP

⇒ PER = (x + y) + (x + y) + 2y

⇒ PER = 2x + 4y …… (2)

From(1)

⇒

⇒

⇒

We need perimeter to be minimum and let us PER as the function of y,

We know that for maxima and minima ,

⇒

⇒

⇒

⇒

⇒ 4y⁴ - 12y²r² = 0

⇒ 4y²(y² - 3r²) = 0

⇒

Differentiating PER again,

⇒

⇒

⇒

⇒

⇒

⇒

⇒ >0(minima)

We got minima at y = r.

Let’s find the value of x,

⇒

⇒ x = r

⇒ PER = 2(r) + 4(r)

⇒ PER = 6r

∴ Thus proved

The perimeter of the rectangle with length L and breadth b is 2(l + b)

Therefore,

2(L + b) = 36

L + b = 18

b = 18 - L

Let the rectangle be rotated about its breadth. Then the resulting cylinder formed will be of radius L and height b.

Volume of cylinder formed V = πL²b = π(18L² - L³)

To find the dimensions that will result in the maximum volume:

L cannot be 0. L is taken as 12 cm.

Therefore b = 24.

At L = 12,

Therefore a maxima exists at L = 12, meaning the volume of the constructed cylinder will be maximum at L = 12 cm.

Let h and r be height and radius of the required cone.

R be the radius of the sphere.

Now, it must be understood that for the cone to have maximum volume, the axis of cone and sphere must be the same.

Let OD = x

In Δ BOD,

AD = AO + OD = 12 + x

The roots of this quadratic equation is - 12 and 4. As - 12 is not possible, we have x = 4.

Therefore, the volume is maximum when the x = 4.

Therefore the height of the cone = R + x = 12 + 4 = 16cm.

Volume of the cylinder = 2156 cm³

Formula for volume: V = πr²h

Therefore,

We need to find the value of r for which the total surface area will be minimum. Hence we rewrite the value of h in terms of r.

Total surface area = 2πr² + 2πrh

Therefore,

Hence,

Therefore, r = 7cm is the answer

The maximum volume cylinder will be carved when the diameter of sphere and the axis of cylinder coincide.

Let h be the height of our cylinder.

r be the radius of the cylinder.

R = 53 radius of the sphere

OL = x

H = 2x

In Δ AOL,

Volume of cylinder V = πr²h

Therefore,

x = - 5 cannot be taken as the length cannot be negative.

At x = 5, we have to check whether maxima exits or not.

Therefore,

At x = 5, is negative. Hence, maxima exits at x = 5.

Therefore, at x = 5

Volume of Cylinder = π(75 - x²)(2x)

Volume = π(75 - 25)(10) = 500π

x,y are positive numbers

x² + y² = r² is the given condition

Therefore,

Let

And,

Given Curve is x² = 4y …… (1)

Let us assume the point on the curve which is nearest to the point (0, 5) be (x, y)

The (x, y) satisfies the relation(1)

Let us find the distance(S) between the points (x, y) and (0, 5)

We know that distance between two points (x₁,y₁) and (x₂,y₂) is .

⇒

⇒

Squaring on both sides we get,

⇒ S² = x² + y² - 10y + 25

From (1)

⇒

⇒

We know that distance is an positive number so, for a minimum distance S, S² will also be minimum

Let us S² as the function of x.

For maxima and minima,

⇒

⇒

⇒

⇒

⇒ x³ - 12x = 0

⇒ x(x² - 12) = 0

On solving we get

⇒

Now differentiating again

⇒

⇒

⇒

At x = 0

⇒

⇒

At x = 2

⇒

⇒

⇒

⇒

At x = - 2

⇒

⇒

⇒

⇒

We get minimum distance at x = ±2

Let find the value of y at these x values

⇒

⇒ y = 3

⇒

⇒ y = 3

∴ The nearest points to the point (0,5) on the curve are (±2,3).

Given Curve is y² = 4x …… (1)

Let us assume the point on the curve which is nearest to the point (2, - 8) be (x, y)

The (x, y) satisfies the relation(1)

Let us find the distance(S) between the points (x, y) and (2, - 8)

We know that distance between two points (x₁,y₁) and (x₂,y₂) is .

⇒

⇒

⇒

Squaring on both sides we get,

⇒ S² = x² + y² - 4x + 16y + 68

From (1)

⇒

⇒

We know that distance is an positive number so, for a minimum distance S, S² will also be minimum

Let us S² as the function of y.

For maxima and minima,

⇒

⇒

⇒

⇒

⇒ y³ - 64 = 0

⇒

On solving we get

⇒

Now differentiating again

⇒

⇒

At y = 4

⇒

⇒

We get minimum distance at y = 4

Let find the value of x at these y values

⇒

⇒ x = 4

∴ The nearest point to the point (2, - 8) on the curve is (4,4).

Given Curve is x² = 8y …… (1)

Let us assume the point on the curve which is nearest to the point (2, 4) be (x, y)

The (x, y) satisfies the relation(1)

Let us find the distance(S) between the points (x, y) and (2, 4)

We know that distance between two points (x₁,y₁) and (x₂,y₂) is .

⇒

⇒

⇒

Squaring on both sides we get,

⇒ S² = x² + y² - 4x - 8y + 20

From (1)

⇒

⇒

We know that distance is an positive number so, for a minimum distance S, S² will also be minimum

Let us S² as the function of x.

For maxima and minima,

⇒

⇒

⇒

⇒

⇒ x³ - 64 = 0

⇒

On solving we get

⇒ x = 4

Now differentiating again

⇒

⇒

At x = 4

⇒

⇒

We get minimum distance at x = 4

Let find the value of y at these x values

⇒

⇒ y = 2

∴ The nearest point to the point (2,4) on the curve is (4,2).

Given Curve is x² = 2y …… (1)

Let us assume the point on the curve which is nearest to the point (0, 5) be (x, y)

The (x, y) satisfies the relation(1)

Let us find the distance(S) between the points (x, y) and (0, 5)

We know that distance between two points (x₁,y₁) and (x₂,y₂) is .

⇒

⇒

Squaring on both sides we get,

⇒ S² = x² + y² - 10y + 25

From (1)

⇒

⇒

We know that distance is an positive number so, for a minimum distance S, S² will also be minimum

Let us S² as the function of x.

For maxima and minima,

⇒

⇒

⇒

⇒ x³ - 8x = 0

⇒ x(x² - 8) = 0

On solving we get

⇒

Now differentiating again

⇒

⇒

At x = 0

⇒

⇒

At x = 2

⇒

⇒

⇒

At x = - 2

⇒

⇒

⇒

We get minimum distance at x = ±2

Let find the value of y at these x values

⇒

⇒ y = 4

⇒

⇒ y = 4

∴ The nearest points to the point (0,5) on the curve are (±2,4).

Given equation of the parabola is y = x² + 7x + 2 and straight line is y = 3x - 3 or 3x - y - 3 = 0

Equation of parabola y = x² + 7x + 2 …… (1)

Let us assume the point on parabola which is closest to the line be (x, y)

We know that distance between the point (x₀, y₀) and the line ax + by + c = 0 is

⇒

Now we find the distance between the line and point on parabola

⇒

From(1)

⇒

⇒

⇒

Let us assume S be the function of x

⇒

⇒

For maxima and minima

⇒

⇒

⇒ 2x + 4 = 0

⇒ 2x = - 4

⇒ x = - 2

Now differentiating again

⇒

⇒

At x = - 2

⇒

We get the minimum distance at x = - 2

Let’s find the value of y at this x value

⇒ y = ( - 2)² + 7( - 2) + 2

⇒ y = 4 - 14 + 2

⇒ y = - 8

∴ The point ( - 2, - 8) is the nearest point on the parabola to the line y = 3x - 3.

Given Curve is y² = 2x …… (1)

Let us assume the point on the curve which is nearest to the point (1, 4) be (x, y)

The (x, y) satisfies the relation(1)

Let us find the distance(S) between the points (x, y) and (1,4)

We know that distance between two points (x₁,y₁) and (x₂,y₂) is .

⇒

⇒

⇒

Squaring on both sides we get,

⇒ S² = x² + y² - 2x - 8y + 17

From (1)

⇒

⇒

We know that distance is an positive number so, for a minimum distance S, S² will also be minimum

Let us S² as the function of y.

For maxima and minima,

⇒

⇒

⇒

⇒ y³ - 8 = 0

⇒

On solving we get

⇒ y = 2

Now differentiating again

⇒

⇒

At y = 2

⇒

⇒

We get minimum distance at y = 2

Let find the value of x at these y values

⇒

⇒ x = 2

∴ The nearest point to the point (2, 2) on the curve is (4,4).

We need to find the maximum value of the slope, meaning we have to double differentiate the value of slope to find the maximum value of slope

Slope (max) = - 3(1) + 6(1) + 2 = 5

Cost Price (CP) for producing x amount

Selling Price (SP) per producing x amount

Selling Price (SP) for producing x amount

Profit P = SP - CP

Therefore,

Maxima exits, therefore we will get maximum profit at x = 10.

Profit P = SP - CP

x = 240

Therefore,

Thus x = 240 is a point of maxima.

Thus for maximum profit, x should be 240.

Let L be the length of the square base and h be the height/depth of the tank.

Expenses of lining implies the cost for lining the entire inner surface area of the tank; a base and four vertical sides.

If we have minimum area to cover, we will have minimum costs incurred.

Internal area of the tank = L² + 4Lh

Volume of the tank = L²h = V

Therefore,

We get two outcomes here.

L = 0,2h

We discard L = 0, as it makes no sense.

So;

L = 2h.

Now to check whether a maxima or a minima exists

Therefore a minima exists for all non zero values of L.

Hence for the tank lining costs to be minimum, h = L/2.

Let,

Width = b

Length L = 2b

Height = h

Cost of material on top and 4 sides = 3 x cost of material at bottom.

Here we do not have individual cost per area of different materials. But we can use the total surface area of individual parts as replacement for the cost.

bx2b + 2(hxb) + 2(hx2b) = 3(bx2b)

2b² + 2hb + 4hb = 6b²

6hb = 4b²

Volume of the box = c = bx2bxh = 2b²h

Therefore,

We want the optimum dimensions of the box. We find this by optimizing the total surface area to be minimum. This is because the majority of the surface costs more.

Surface area(total) = S = 2(bh + 2bh + 2b²) = 2(2b² + 3bh)

A minima exists at

Therefore:

And Length = 2b

Let us assume radius of sphere be ‘r’ and length of side of cube is ‘l’

We know that,

⇒ Surface area of sphere = 4r²

⇒ Surface area of cube = 6l²

According the problem, the sum of surface areas of a sphere and cube is known. Let us assume the sum be S

⇒ S = 4r² + 6l² …… (1)

We also know that,

⇒ Volume of sphere =

⇒ Volume of cube = l³

We need the sum of volumes to be least. Let us assume the sum of volumes be V

⇒

From (1)

⇒

We assume V as a function of r.

For maxima and minima,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Differentiating V again

⇒

⇒

⇒

⇒

At r = 0

⇒

⇒

At

⇒

⇒

⇒

⇒

We get the sum of values least for .

We know that diameter(d) is twice of radius. So,

⇒

⇒ d = l

∴ Thus proved.

Let ‘h’ be the height ‘or’ length of half cylinder, ‘r’ be the radius of half cylinder and ‘d’ be the diameter.

We know that,

⇒ Volume of half cylinder (V) =

⇒ …… (1)

Now we find the Total surface area (TSA) of the half cylinder,

⇒ TSA = Lateral surface area of the half cylinder + Area of two semi - circular ends + Area of the rectangular base

⇒

From (1)

⇒

⇒

We need total surface area to be minimum and let us take the TSA as the function of r,

For maxima and minima,

⇒

⇒

⇒

⇒

⇒

⇒

Differentiating TSA again,

⇒

⇒

⇒

⇒

At

⇒

⇒

⇒ >0(Minima)

We have got Total surface area minimum for

We know that diameter is twice of radius

⇒

⇒

⇒

∴ Thus proved.

Let us assume ABCD be the cross section of beam that need to cut from the circular log of radius ‘a’.

Let us assume ‘b’ be the depth of the rectangle, ‘l’ be the length of the rectangle and ‘d’ be the diameter of circular log.

⇒ d = 2a …… (1)

According to the problem, strength of the beam is given by

⇒ S = lb² …… (2)

From the ΔABC,

⇒ d² = l² + b²

From (1)

⇒ (2a)² = l² + b²

⇒ b² = 4a² - l² …… (3)

From(2) and (3)

⇒ S = l(4a² - l²)

⇒ S = 4a²l - l³

We need strength of the beam to be maximum, let us take S as a function of l

For maxima and minima,

⇒

⇒

⇒

⇒ 3l² = 4a²

⇒

⇒

Differentiating S again,

⇒

⇒

At ,

⇒

⇒ <0(Maxima)

We get maximum strength for length , lets find the depth ‘b’ for this l:

⇒

⇒

⇒

⇒

Let’s find the strength of beam for and

⇒

⇒

The dimensions of beam of maximum strength is .

Let us the slope of the line passing through the point P(1,4) be m.

We know that equation of a straight line passing through the point (x₁,y₁) and having slope m is given by:

⇒ y - y₁ = m(x - x₁)

The equation of the straight line is:

⇒ y - 4 = m(x - 1)

⇒ mx - y = m - 4

⇒

⇒

This resembles the standard form , where a is x - intercept and b is y - intercept.

Here x - intercept and b = 4 - m

According to the problem, we need sum of intercepts to be minimum,

Let us take the sum of intercepts to be S,

⇒ S = a + b

⇒

⇒

Let us assume S is the function of m,

We know that for maxima and minima,

⇒

⇒

⇒

⇒ 4 - m² = 0 (∵ m²>0)

⇒ m = ±2

Differentiating S again

⇒

⇒

At m = - 2

⇒

⇒

⇒ >0(Minima)

At m = + 2

⇒

⇒

⇒ <0(Maxima)

We have got minima for m = - 2

Using this value we find the sum of intercepts:

⇒

⇒

⇒ S_min = 3 + 6

⇒ S_min = 9

∴ The least value of sum of intercepts is 9.

Let ABCD be the total page and A’B’C’D’ be the area in which the matter is printed.

Let ‘l’ and ‘b’ be the length and breadth of the total page.

We know that area of the rectangle is l×b

From the problem,

⇒ lb = 150cm² …… (1)

⇒ PQ + RS = 3cm

⇒ WX + YZ = 2cm

Let ‘l’’ and ‘b’’ be the length and breadth of the area in which the matter is printed,

From the figure,

⇒ l’ = (l - 2)cm

⇒ b’ = (b - 3)cm

⇒ Area of the printed matter (A) = (l - 2)(b - 3)cm²

From (1)

⇒

⇒

⇒

We need Area of the printed matter maximum and let us take A as the function of l

We know that for maxima and minima,

⇒

⇒

⇒

⇒ (∵ l²>0)

⇒

⇒ l = 10cm (since length is an positive quantity)

Differentiating A again,

⇒

⇒

At l = 10

⇒

⇒ <0(Maxima)

We get the area of the printed matter maximum for l = 10cm

Let’s find the corresponding breadth using eq(1)

⇒ 10b = 150

⇒ b = 15cm

∴ The dimensions required for the printed area to be maximum is l = 10cm and b = 15cm.

Given:

The Distance(S) covered by a particle in time t is given by

⇒ S = t⁵ - 40t³ + 30t² + 80t - 250

We know that acceleration of a particle is given by .

⇒

⇒

⇒

⇒

We need acceleration to be minimum,

We know that for maxima and minima,

⇒

⇒

⇒

⇒ t² = 4

⇒ t = 2 (∵ Time cannot be negative)

Differentiating ‘a’ again,

⇒

⇒

At t = 2

⇒

⇒ >0(Minima)

We get minimum for t = 2 sec,

The corresponding acceleration at t = 2 sec is,

⇒ a = 20(2)³ - 240(2) + 60

⇒ a = 160 - 480 + 60

⇒ a = - 260

∴ The minimum acceleration is - 260.

Given:

The distance covered by a particle at time ‘t’ is given by,

⇒

We know that velocity of a particle is given by and acceleration of a particle is given by .

⇒

⇒

We need velocity to be maximum,

We know that for maxima and minima,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Differentiating ‘v’ again,

⇒

⇒

At

⇒

⇒

⇒ >0(Minima)

At

⇒

⇒

⇒ <0(Maxima)

We get the velocity maximum at

Now, we find the acceleration:

⇒

⇒ a = 3t² - 12t + 8

We need acceleration to be minimum,

We know that for maxima and minima,

⇒

⇒

⇒

⇒ t = 2

Differentiating ‘a’ again,

⇒

⇒

At t = 2

⇒ >0(Minima)

We get minimum for t = 2 ,

∴ we get maximum velocity at and minimum acceleration at t = 2

.

Let

therefore,.

Differentiating w.r.t x

so,

Now, lets put y’=0

soOR1-log x) =0

therefore,x=1/e and x=0

Hence by second derivative test

f’’(1/e)<0 so it’s a point of maximum

and maximum value is.

let f(x)=ax+(b/x)-c;

x>0 where a,b>0

f’(x)=0

f’(x)=0

_logx-1=0

_⇒x=e

_{for second derivative we find f’(x)}

Hence by second derivative test

f’’(x)>0 so it’s a point of minimum.

therefore,

x=e is a point of minimum

so minimum value is f(e)=e

In such type of questions

find both the maximum and minimum value to compare the options.

so first,

so,

put f’(x)=0;

⇒ x=±1

Hence by second derivative test

f’’(x)>0 or f”(x)<0

so it’s a point of minimum or maximum respecctively.

f”(-1)=-2<0;

f”(1)=2>0

so x=1 is a point of minimum

and x=-1 is a point of maximum

f(1)=2 is minimum value.

f(-1)=-2 is maximum value.

therefore,

maximum value<minimum value.

f(x)=x³+3x²-9x+2

f’(x)=3x²+6x-9

f’(x)=0

3x²+6x-9=0

by solving the quadratic we get roots as follows

x=-3 and x=1

Hence by second derivative test

f’’(x)>0 or f”(x)<0 so it’s a point of minimum or maximum respecctively.

f”(x)=6x+6

f”(-3)=6(-3)+6

=-12<0

so At x=-3 maximum value.

f”(1)=6+6

=12>0

so At x=1 minimum value.

Option=(B)

f(x)=x⁴-x²-2x+6

f’(x)=4x³-2x-2

so here put f’(x)=0

4x³-2x-2=0

2(x-1)(2x²+2x+1)=0

x=1 and other roots are complex.

so we will consider x=1

Hence by second derivative test

f’’(x)>0 so it’s a point of minimum.

f”(x)=12x²-1

f” (1) =12(1)-2

=10>0

At x=1 minimum

So, f (1) =1-1-2(1) +6

=4

Option(B)

Let the function be according to the question,

f(x)=x-x²

Here,it is asked “greatest possible quantity” so indirectly they are asking for the maximum value at which point x=?

so,f’(x)=1-2x

f’(x)=0

=>1-2x=0

x= �

f”(x)=-2<0

So for x= � maximum value of the function.

Option(A).

f(x)=(x-a)² +(x-b)² +(x - c)²

f'(x)=2[x – a + x – b + x - c]

f'(x)=2[3x-a-b-c]

f’(x)=0

2[3x-a-b-c] =0

3x-a-b-c=0

3x=a + b + c

Hence by second derivative test

f’’(x)>0 so it’s a point of minimum.

f” (x)=2(3(1))

=6>0

so point of minimum.

Option(A).

Let the two non zero numbers be x and y.

so,x+y=8

y=8-x

and,

put y=8-x in f(x,y)

therefore it will become a function in

f’(x)=0

⇒ x=8-x

x=4

Hence by second derivative test

f’’(x)>0 so it’s a point of minimum.

Hence at x=4 minimum value

Option(B)

f(x)=(x-1)²+(x-2)²+(x-3)²+(x-4)²+(x-5)²

f’(x)=2[5x-15]

f’(x)=0;x=3

Hence by second derivative test

f’’(x)>0 so it’s a point of minimum.

f”(x)=1>0 so At x=3 minimum value.Option(C)

f’(x)=0

⇒6 cos 3x – 9 sin 3x = 0

⇒6 cos 3x = 9 sin 3x

So, at _{neither is maximum nor minimum}

_Option(D)

f(x)=

f’(x)=

f’(x)=0

⇒2x+1=0

⇒ x= - �

At extreme points,

f(0)=0

f(1)=1+1+1>0

so,x=1 is a least value.

Option(C)

f(x)=x³-18x²+96x

f(x)=3x²-36x+96

f’(x)=0

3x²-36x+96=0

x²-12x+32=0

x²-8x-4x+32=0

x(x-8)-4(x-8)=0

Now we have f(0)=0,f(4)=160,f(9)=135

Hence in the given interval least value is f(0)=0

Option(D)

=0

⇒4-x〗^2=0

x= but [-1,1]

=>f(-1)=-1/6 Option(C)

If we consider the above figure as for getting it,their will be many points on the curve.

If the normal to the point B passes through the point(2,1) then point B will be the point having nearest distance from point (2,1).

Let B(x,y)

=

slope at the point B is (2/y) and normal’s slope will be m=(–y/2) so by point slope formula.

⇒ (y-y1)=m(x-x1);(x1=2,y1=1)

⇒ (y-1)=(-y/2)(x-2)

⇒2y-2=-xy+2y

⇒xy=2;

⇒from above to equations

⇒y=2 and x=1

so the nearest point is (1,2).

Option(B)

x+y=8 ⇒ y=8-x

xy=x(8-x)

Let f(x)=8x-x²

Differentiating f(x) with respect to x, we get

f’(x)=8-2x

Differentiating f’(x) with respect to x, we get

f’’(x)=-2<0

For maxima at x=c, f’(c)=0 and f’’(c)<0

f’(x)=0 ⇒ x=4

Also f’’(4)=-2<0

Hence, x=4 is a point of maxima for f(x) and f(4)=16 is the maximum value of f(x).

f(x) = x³ – 6x² + 9x, x∈[0,6]

Differentiating f(x) with respect to x, we get

f’(x)= 3x² - 12x + 9=3(x-3)(x-1)

For extreme points, f’(x)=0 ⇒ x=1 or x=3

For least and greatest value of f(x) in [0,6], we will have to check at extreme points as well as interval extremes

f(1)=4

f(3)=0

f(0)=0

f(6)=54

Hence the least value of f(x) in [0,6] is 0 and it’s greatest value is 54.

Differentiating f(x) with respect to x, we get

Differentiating f’(x) with respect to x, we get

For maxima at x=c, f’(c)=0 and f’’(c)<0

f’(x)=0⇒ or

and >0

Hence, is a point of maxima for f(x).

In the figure, ∆PQR represents the 2D view of the cone and the circle represents the sphere. PA is perpendicular to QR and PS is diameter of circle. C will lie on PA due to symmetry.

Let the radius and height of cone be r and h and the radius of sphere be R. Also, the semi vertical angle of cone is α.

In ∆PAR

-(1)

∠APR=α

∠PAR=90°

Hence, ∠ PRA=180° -90° -α =90° -α

Also ∠PRS=90° (Angle in a semicircle)

Hence ∠ARS=∠PRS-∠PRA=α

In ∆RAS

AS=PS-PA=2R-h

–(2)

From (1) and (2), we get

r²=2Rh-h²

The volume of cone will be

Differentiating V with respect to h, we get

Differentiating V’ with respect to h, we get

For maxima at h=c, V’(c)=0 and v’’(c)<0

V’=0 ⇒

Hence, the ratio of height of cone to diameter of sphere is .

Differentiating f(x) with respect to x, we get

Differentiating f’(x) with respect to x, we get

for minima at x=c, f’(c)=0 and f’’(c)>0

f’(x)=0 ⇒ x³=125 or x=5

f’’(5)=7>0

Hence, x=5 is a point of minima for f(x) and f(5)=75 is the minimum value of (x).

Differentiating f(x) with respect to x, we get

Also, differentiating f’(x) with respect to x, we get

For maxima at x=c, f’(c)=0 and f’’(c)<0

⇒ or (Since x>0)

f’’(1)=2>0

Since f’’(1)>0 therefore x=1 is not a point of maxima and hence no maximum value of f(x) exists for x>0.

xy=1, x>0, y>0

⇒

x+y=

Let

Differentiating f(x) with respect to x, we get

Also, differentiating f’(x) with respect to x, we get

For minima at x=c, f’(c)=0 and f’’(c)<0

⇒ or (Since x>0)

f’’(1)=2>0

Hence, x=1 is a point of minima for f(x) and f(1)=2 is the minimum value of f(x) for x>0.

f(x)=1+2sinx+3cos²x,

Differentiating f(x) with respect to x, we get

f’(x)=2cosx-6cosxsinx

Also, differentiating f’(x) with respect to x, we get

f’’(x)=-2sinx-6cos²x+6sin²x=-2sinx+12sin²x-6 (Since sin²x+cos²x=1)

For extreme points, f’(x)=0

⇒ 2cosx(1-3sinx)=0 or or

and

Hence, is a point of minima and is a point of maxima.

f(x) = 2x³ – 15x² + 36x + 4

Differentiating f(x) with respect to x, we get

f’(x)= 6x² - 30x + 36=6(x-2)(x-3)

Differentiating f’(x) with respect to x, we get

f’’(x)=12x – 30

for maxima at x=c, f’(c)=0 and f’’(c)<0

f’(x)=0 ⇒ x=2 or x=3

f’’(2)=-6<0 and f’’(3)=6>0

Hence x=2 is a point of maxima.

Differentiating f(x) with respect to x, we get

Since 4-x²>0 ∀ x∈[-1,1] and (4+x+x²)²>0 ∀ x∈R

Therefore, f’(x)>0 ∀ x∈[-1,1]

Hence, f(x) is increasing in [-1,1] and therefore the maximum value of f(x) occurs at x=1 and

f(x) = 2x³ – 3x² – 12x + 5, x∈[-2,4]

Differentiating f(x) with respect to x, we get

f’(x)= 6x² – 6x – 12=6(x+1)(x-2)

Differentiating f’(x) with respect to x, we get

f’’(x)=12x-6

For maxima at x=c, f’(c)=0 and f’’(c)<0

f’(x)=0 ⇒ x=-1 or 2

f’’(-1)=-18<0 and f’’(2)=18>0

Hence, x=-1 is the point of local maxima.

Differentiating f(x) with respect to x, we get

Differentiating f’(x) with respect to x, we get

for minima at x=c, f’(c)=0 and f’’(c)>0

f’(x)=0 ⇒

Hence, is a point of minima for f(x) and is the minimum value of f(x).

f(x) = 2x³ – 21x² + 36x – 20

Differentiating f(x) with respect to x, we get

f’(x)= 6x² - 42x + 36=6(x-1)(x-6)

Differentiating f’(x) with respect to x, we get

f’’(x)=12x-42

for minima at x=c, f’(c)=0 and f’’(c)>0

f’(x)=0 ⇒ x=1 or x=6

f’’(1)=-30<0 and f’’(6)=30>0

Hence, x=6 is the point of minima for f(x) and f(6)=-128 is the local minimum value of f(x).

achieves it’s maximum value when g(x)=4x²+2x+1 achieves it’s minimum value.

Differentiating g(x) with respect to x, we get

g’(x)=8x+2

Differentiating g’(x) with respect to x, we get

g’’(x)=8

For minima at x=c, g’(c)=0 and g’’(c)>0

g’(x)=0 ⇒