# Class 12 RD Sharma Solutions – Chapter 18 Maxima and Minima – Exercise 18.5 | Set 2

### Question 16. A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 meters, find the dimensions of the rectangle that will produce the largest area of the window.

Solution:

According to the question

Let us assume l be the length of the rectangle and b be the breadth of the rectangle

The perimeter of the window = 12 m

â‡’ (l + 2b) + (l + l) = 12

â‡’ 3l + 2b = 12     ……(i)

Mow we find the area of the window (A) = Area of the rectangle + Area of the equilateral â–³

A = l (12 – 3l / 2) + âˆš3/4 l2

On differentiating w.r.t. l, we get

dA/dl = 6 – 3l + (âˆš3/2)l = 6 – âˆš3(âˆš3 – 1/2)l

For maxima and minima,

Put dA/dl = 0

â‡’ 6 – âˆš3(âˆš3 – 1/2)l = 0

â‡’ l = 6/{âˆš3(âˆš3 – 1/2) } = 12- (6 – âˆš3)

Now, d2A/dl2 = -âˆš3(âˆš3 – 1/2) = -3 + âˆš3/2

So, l = 12/(6 – âˆš3) is the point of local maxima

So, When l = 12/(6 – âˆš3), the area of the window is maximum

From eq(i), we get

b = (12 – 3l)/2 = [12 – 3{12/(6 – âˆš3)}]/2 = (24 – 6âˆš3)/(6 – âˆš3)

### Question 17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/âˆš3. Also, find the maximum volume.

Solution:

According to the question

R be the radius of the sphere

So, let us assume that r and h be the radius and the height of the cylinder

So, according to the image

h = 2 âˆš(R2 – r2)

Now we find the volume of the cylinder is

V = Ï€r2h = 2Ï€r2âˆš(R2 – r2)

On differentiating w.r.t. r, we get

dV/dr = 4Ï€r âˆš(R2 – r2) +

= 4Ï€r âˆš(R2 – r2) –

For maxima and minima,

Put dV/dr = 0

4Ï€rR2 – 6Ï€r3 = 0

r2 = 2R2/3

Now, again differentiating w.r.t. r, we get

d2V/dr2

So, at r2 = 2R2/3, d2V/dr2 < 0

Hence, the volume is the maximum when r2 = 2R2/3

so, the height of the cylinder = 2âˆš(R2 – 2R2/3) = 2âˆš(R2/3) = 2R/âˆš3

Hence proved

### Question 18. A rectangle is inscribed in a semicircle of radius r with one of its sides on diameter of semicircle. Find the dimensions of the rectangle so that its area is maximum Find also the area.

Solution:

Let us assume EFGH be a rectangle inscribed in a semi-circle. So, r be the radius of the semicircle.

And l and b are the length and width of rectangle.

Now In â–³OHE,

HE2 = OE2 – OH2

HE = b =   …..(i)

Now we find the area of the EFGH rectangle

A = lb = l Ã—

A = 1/2 l âˆš(4r2 – l2)

On differentiating w.r.t. l, we get

dA/dl = 1/2 \sqrt{4r^2-l^2}-\frac{l^2}{\sqrt{4r^2-l^2}}

= 1/2 \frac{4r^2-l^2-l^2}{\sqrt{4r^2-l^2}}

For maxima and minima,

Put dA/dl = 0

â‡’

â‡’ l = Â±âˆš2r

As we know that l can’t be negative so l â‰  -âˆš2r

So, when l = âˆš2r, d2A/dl2 < 0

Hence, the area of the rectangle is maximum when l = âˆš2r

Now put the value of l = âˆš2r in eq(i), we get

Now we find the area of rectangle = lb

= âˆš2r Ã— r/âˆš2

= r2

### Question 19. Prove that a conical tent of given capacity will require the least amount of canvas when the height is âˆš2 times the radius of the base.

Solution:

Let us assume that the radius and height of the cone is r and h.

So, the volume of the cone is

V = 1/3 Ï€r2h

h = 3V/r2  ……(i)

And the surface area of the cone is

A = Ï€rl

Here, l is the slant height = âˆš(r2 + h2)

= Ï€râˆš(r2 + h2)

On differentiating w.r.t. r, we get

dA/dr =

=

For maxima and minima,

Put dA/dr = 0

2Ï€2r6 = 9V2

r6 = 9V2/2Ï€2

When, r6 = 9V2/2Ï€2, d2S/dr2 > 0

Hence, the surface area of the cone is the least when r6 = 6V2/2Ï€2

Now put r6 = 9V2/2Ï€2 in eq(i), we get

h = 3V/Ï€r2 = 3/Ï€r2(2Ï€2r6/9)1/2 = (3/Ï€r2)(âˆš2Ï€r3/3) = âˆš2r

Hence Proved

### Question 20. Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to 2/3 of the diameter of the sphere.

Solution:

Let us assume R be the radius of the sphere

So, from the figure, we get OD = x and AO = OB = R

BD = âˆš(R2 – x2) and AD = (R + x)

Now,

The volume of the cone is

V = 1/3 Ï€r2h

= 1/3 Ï€ (R2 – x2) Ã— (R + x)

On differentiating w.r.t. x, we get

dV/dx = Ï€/3 [-2x (R + x) + R2 – x2]

= Ï€/3 [R2 – 2xR – 3x2]

For maximum and minimum

Put dv/dx = 0

â‡’ Ï€/3 [R2– 2xR – 3x2] = 0

â‡’ Ï€/3 [(R – 3x) (R + x)] = 0

â‡’ R  – 3x = 0 or x = -R

Here, x = -R is not possible because -r will make the altitude 0

â‡’ x = R/3

Now,

d2V/dx2 = Ï€/3[-2R – 6x]

So, when x = R/3, d2v/dx2 = Ï€/3[-2R – 2R] = -4Ï€R/3 < 0

So, x = R/3 is the point of local maxima.

Hence, the volume is maximum when x = R/3

So, the altitude AD = (R + x) = (R + R/3) = 4R/3 = 2/3d

Here, d is the diameter of sphere.

### Question 21. Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is cot-1 (âˆš2).

Solution:

Let us assume h, r and Î¸ be the height, radius and semi vertical angle of the right-angled triangle.

So, the volume of the cone (V) = 1/3 Ï€r2h

â‡’ h = 3V/Ï€r2

Slant height of the cone (l) = âˆš(r2 + h2)

l =

And the curved surface area of the cone is

A = Ï€rl

A = Ï€r

A =

On differentiating w.r.t. r, we get

dA/dr =

For maximum and minimum

Put dA/dr = 0

= 0

â‡’ 2Ï€2r6 – 9V2 = 0

â‡’ V2 = 2Ï€2r6/9

â‡’ V = âˆš2Ï€2r6/9

â‡’ V = Ï€r3âˆš2/3

or

r = (3V/Ï€âˆš2)1/3

h/r = âˆš2

cotÎ¸ = âˆš2

Semi-vertical angle, Î¸ = cot-1âˆš2

Also, when r < (3V/Ï€âˆš2)1/3, dA/dr < 0

When r > (3V/Ï€âˆš2)1/3, dA/dr > 0

Hence, the curved surface for r = (3V/Ï€âˆš2)1/3 is least.

### Question 22 An isosceles triangle of vertical angle 2Î¸ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when Î¸ = Ï€/6.

Solution:

Let us considered ABC is an isosceles triangle such that AB = AC

and the vertical angleâˆ BAC = 2Î¸

Radius of the circle = a

Now, draw AM perpendicular to BC.

From the figure we conclude that in â–³ABC is an isosceles triangle

the circumcenter of the circle lies on the perpendicular from A to BC and

O be the circumcenter of the circle

So, âˆ BOC = 2 Ã— 2Î¸ = 4Î¸

and âˆ COM = 2Î¸   [Since â–³OMB and â–³OMC are congruent triangles]

OA = OB = OC =a       [Radius of the circle]

In â–³OMC,

CM = asin2Î¸ and OM = acos2Î¸

BC = 2CM  [Perpendicular from the centre bisects the chord]

BC = 2asin2Î¸                             …..(i)

In â–³ABC,

AM = AO + OM

AM = a + acos2Î¸                          …..(ii)

Now the area of â–³ABC is,

A = 1/2 Ã— BC Ã— AM

= 1/2 Ã— 2asin2Î¸ Ã— (a + acos2Î¸)   ……(iii)

On differentiating w.r.t. Î¸, we get

dA/dÎ¸ = a2(2cos2Î¸ + 1/2 Ã— 4cos4Î¸)

dA/dÎ¸ = 2a2 (cos2Î¸ + cos4Î¸)

Again differentiating w.r.t. Î¸, we get

d2A/dÎ¸2 = 2a2(-2sin2Î¸ – 4sin4Î¸)

For maximum and minimum

Put dA/dÎ¸ = 0

2a2(cos2Î¸ + cos4Î¸) = 0

cos2Î¸ + cos4Î¸ = 0

cos2Î¸ + 2cos22Î¸ – 1 = 0

(2cos2Î¸ – 1)(2cos2Î¸ + 1) = 0

cos2Î¸ = 1/2 or cos2Î¸ = -1

2Î¸ = Ï€/3 or 2Î¸ = Ï€

Î¸ = Ï€/6 or Î¸ = Ï€/2

When Î¸ = Ï€/2, it will not form a triangle.

When Î¸ = Ï€/6, d2A/dÎ¸2 < 0

Hence, the area of the triangle is maximum when Î¸ = Ï€/6

### Question 23. Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides.

Solution:

Let us assume l, b, and V be the length, breadth, and volume of the rectangle.

The perimeter of the rectangle is 36cm

2(l + b) = 36

l + b = 18

l = 18 – b ……(i)

V = Ï€l2b

V = Ï€(18 – b )2b

V = Ï€(324 + b2 – 36b)b

V = Ï€(324b + b3 – 36b2)

On differentiating w.r.t. b, we get

dV/db = Ï€(324 + 3b2 – 72b)

Again differentiating w.r.t. b, we get

d2V/db2 = Ï€(6b – 72)

For maximum and minimum

Put dV/db = 0

Ï€(324b + b3 – 36b2) = 0

(b – 6)(b – 18) = 0

b = 6, 18

When b = 6, d2V/db2 = -36Ï€ < 0

When b = 18, d2V/db2 = 36Ï€ > 0

So, at b = 6 is the maxima

Hence, the volume is maximum when b = 6

Now put the value of b in eq(i), we get

l = 18 – 6

l = 12

Hence, the dimension of rectangle are 12 cm and 6 cm.

### Question 24. Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.

Solution:

Let us assume r and h be the radius of the base of cone and height of the cone.

From the figure OD = x, and R = 12, BD = r

In â–³BOD,

BD = âˆš(R2 – x2)

= âˆš(144 – x2)

= (144 – x2)

and AD = AO + OD

= R + x = 12 + x

The volume of cone is

V = 1/3 Ï€r2h

= 1/3 Ï€ BD2 Ã— AD

= 1/3 Ï€(144 – x2)(12 + x)

= 1/3 Ï€(1728 + 144x – 12x2 – x3)

On differentiating w.r.t. x, we get

dV/dx = 1/3Ï€ (144 – 24x – 3x2)

For maximum and minimum

Put dV/dx = 0

â‡’ 1/3 Ï€(144 – 24x – 3x2) = 0

â‡’ x = -12, 4

Here x = -12 is not possible

So, x = 4

Now,

d2V/dx2 = Ï€/3(-24 – 6x)

At x = 4, d2v/dx2 = -2Ï€(4 + x) = -2Ï€ Ã— 8 = -16Ï€ < 0

So, x = 4 is point of local maxima.

Hence, the height of cone of maximum volume = R + x

= 12 + 4 = 16 cm

### Question 25. A closed cylinder has volume 2156 cm3. What will be the radius of its base so that its total surface area is minimum?

Solution:

Given that the volume of the closed cylinder (V) = 2156 cm3

Let us assume r and h be the radius and the height of the cylinder.

So, the volume of the cylinder is

V= Ï€r2h = 2156              …..(i)

and the total surface area is

A = 2Ï€rh + 2Ï€r2

A = 2Ï€r (h + r)        …..(ii)

So, from eq (i) and (ii)

A = (2156 Ã— 2)/r + 2Ï€r2

On differentiating w.r.t. r, we get

dA/dr = – 4312/4Ï€ + 4Ï€r

For maximum and minimum

Put dA/dr = 0

â‡’ (-4312 + 4Ï€r3)/r2 = 0

â‡’ r3 = 4312/4Ï€

â‡’ r = 7

At r = 7, d2s/dr2 = (8624/r3 + 4Ï€) > 0

So, r = 7 is the point of local minima

Hence, the total surf surface area of closed cylinder will be minimum when r = 7 cm.

### Question 26. Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius 5âˆš3 cm is 500Ï€ cm3.

Solution:

Let r and h be the radius and height of the cylinder.

Given that R be the radius of the sphere = 5âˆš3

So, from the figure LM = h, OL = x

So, h = 2x

Now, In â–³AOL,

AL = âˆš(AO2 – OL2)

= âˆš(75 – x2)

As we know that the volume of cylinder is

V = Ï€r2h

= Ï€AL2 Ã— ML

= Ï€(75 – x2) Ã— 2x

On differentiating w.r.t. x, we get

dV/dx = Ï€[150 – 6x2]

For maximum and minimum

Put dV/dx = 0

â‡’ Ï€[150 – 6x2] = 0

â‡’ x = 5 cm

Also, d2v/dx2 = -12Ï€x

At x = 5, d2v/dx2 = -60Ï€x < 0

So, x = 5 is point of local maxima.

Hence, the volume is maximum when x = 5

So, the maximum volume of cylinder is

= Ï€(75 – 25) Ã— 10 = 500Ï€ cm3

### Question 27. Show that among all positive numbers x and y with x2 + y2 = r2, the sum x + y is largest when x = y = r/âˆš2.

Solution:

Let us considered the two positive numbers are x and y with

x2+y2 = r2     ……(i)

Let S be the sum of two positive numbers

S = x + y   …..(ii)

= x + âˆš(r2 – x2)       [From eq(ii)]

On differentiating w.r.t. x, we get

dS/dx = 1 –  x/âˆš(r2 – x2)

For maximum and minimum

Put dS/dx = 0

â‡’ 1 – x/âˆš(r2 – x2) = 0

â‡’ x = âˆš(r2 – x2)

â‡’ 2x2 = r2

â‡’ x = r/âˆš2, -r/âˆš2

According to the question x and y are the positive numbers

So, x â‰  -r/âˆš2

Now, d2S/dx2

At, x = r/âˆš2, d2S/dx2 < 0

So, x = r/âˆš2 is point of local maxima.

Hence, the sum is largest when x = y = r/âˆš2

### Question 28. Determine the points on the curve x2 = 4y which are nearest to the point (0, 5).

Solution:

The given equation of parabola is

x2 = 4y  …….(i)

Let us considered P(x, y) be the nearest point of the given parabola from the point A (0, 5)

Let Q be the square of the distance of P from A

Q = x2 + (y – 5)2        …..(ii)

Q = 4y + (y – 5)2

On differentiating w.r.t. y, we get

â‡’ dQ/dy = 4 + 2(y – 5)

For maximum and minimum

Put dQ/dy = 0

â‡’ 4 + 2(y – 5) = 0

â‡’ 2y = 6

â‡’ y = 3

From eq(i), we get

x2 = 12

x = 2âˆš3

â‡’ P = (2âˆš3, 3) and p’ = (-2âˆš3, 3)

Now,

d2Q/dy2 = 2 > 0

So, P and P’ are the point of local minima.

Hence, the nearest points are P(2âˆš3, 3) and P”(2âˆš3, 3).

### Question 29. Find the point on the curve y2 = 4x which is nearest to the point (2, -8).

Solution:

The given equation of the curve is

y2 = 4x ….(1)

Let us assume P(x, y) be a point on the given curve and

Q be the square of the distance between A(2,-8) and P.

So, Q = (x – 2)2 + (y + 8)2   …….(ii)

= (y2/4 – 2)2 + (y + 8)2

On differentiating w.r.t. y, we get

dQ/dy = 2(y2/4 – 2) Ã— y/2 + 2(y + 8)

= (y3 – 8y)/4 + 2y + 16

= y3/4 + 16

For maximum and minimum

Put dQ/dy = 0

â‡’ y3/4 + 16 = 0

â‡’ y = -4

Now,

d2Q/dy2 = 3y2/4

At y = -4, d2S/dy2 = 12 > 0

Ao, y = -4 is the point of local minima

Now put the value of y in eq(i), we get

x = y2/4 = 4

Hence, the point is(4, -4) which is nearest to (2, -8).

### Question 30. Find the point on the curve x2 = 8y which is nearest to the point (2, 4).

Solution:

The given equation of the curve is

x2 = 8y   ….(1)

Let P(x, y) be a point on the given curve, and

Q be the square of the distance between P and A(2, 4).

Q = (x – 2)2 + (y – 4)2   ……..(ii)

= (x – 2)2 + (x2/8 – 4)2

On differentiating w.r.t. x, we get

dQ/dx = 2(x – 2) + 2(x2/8 – 4) Ã— 2x/8

= 2(x – 2) + (x2 – 32)x/16

Also, d2Q/dx2 = 2 + 1/16[x2 – 32 + 2x2]

= 2 + 1/16[3x2 – 32]

For maxima and minima,

dQ/dx = 0

â‡’ 2(x – 2) + x(x2 – 32)/16 = 0

â‡’ 32x – 64 + x3 -32x = 0

â‡’ x3 – 64 = 0

â‡’ x = 4

At x = 4, d2Q/dx2 = 2 + 1/16[16 Ã— 3 – 32] = 2 + 1 = 3 > 0

So, x = 4 is point of local minima

Now put the value of x in eq(1), we get

y = x2/8 = 2

So, the P(4, 2) is the nearest point to (2, 4)

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