# 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 l

^{2}

On differentiating w.r.t. l, we getdA/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, d

^{2}A/dl^{2 }= -âˆš3(âˆš3 – 1/2) = -3 + âˆš3/2So, 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(R

^{2 }– r^{2})Now we find the volume of the cylinder is

V = Ï€r

^{2}h = 2Ï€r^{2}âˆš(R^{2 }– r^{2})

On differentiating w.r.t. r, we getdV/dr = 4Ï€r âˆš(R

^{2 }– r^{2}) += 4Ï€r âˆš(R

^{2 }– r^{2}) –=

=

For maxima and minima,

Put dV/dr = 0

4Ï€rR

^{2}– 6Ï€r^{3}= 0r

^{2}= 2R^{2}/3Now, again

differentiating w.r.t. r, we getd

^{2}V/dr^{2}==

=

So, at r

^{2}= 2R^{2}/3, d^{2}V/dr^{2}< 0Hence, the volume is the maximum when r

^{2}= 2R^{2}/3so, the height of the cylinder = 2âˆš(R

^{2 }– 2R^{2}/3) = 2âˆš(R^{2}/3) = 2R/âˆš3Hence 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,

HE

^{2}= OE^{2}– OH^{2}HE = b = …..(i)

Now we find the area of the EFGH rectangle

A = lb = l Ã—

A = 1/2 l âˆš(4r

^{2}– l^{2})

On differentiating w.r.t. l, we getdA/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, d

^{2}A/dl^{2 }< 0Hence, 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

= r

^{2}

### 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€r

^{2}hh = 3V/r

^{2}……(i)And the surface area of the cone is

A = Ï€rl

Here, l is the slant height = âˆš(r

^{2 }+ h^{2})= Ï€râˆš(r

^{2 }+ h^{2})=

On differentiating w.r.t. r, we getdA/dr =

=

For maxima and minima,

Put dA/dr = 0

2Ï€

^{2}r^{6}= 9V^{2}r

^{6}= 9V^{2}/2Ï€^{2}When, r

^{6}= 9V^{2}/2Ï€^{2}, d^{2}S/dr^{2}> 0Hence, the surface area of the cone is the least when r

^{6}= 6V^{2}/2Ï€^{2}Now put r

^{6}= 9V^{2}/2Ï€^{2 }in eq(i), we geth = 3V/Ï€r

^{2}= 3/Ï€r^{2}(2Ï€^{2}r^{6}/9)^{1/2 }= (3/Ï€r^{2})(âˆš2Ï€r^{3}/3) = âˆš2rHence 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(R

^{2 }– x^{2}) and AD = (R + x)Now,

The volume of the cone is

V = 1/3 Ï€r

^{2}h= 1/3 Ï€BD

^{2}Ã— AD= 1/3 Ï€ (R

^{2 }– x^{2}) Ã— (R + x)

On differentiating w.r.t. x, we getdV/dx = Ï€/3 [-2x (R + x) + R

^{2 }– x^{2}]= Ï€/3 [R

^{2 }– 2xR – 3x^{2}]For maximum and minimum

Put dv/dx = 0

â‡’ Ï€/3 [R

^{2}– 2xR – 3x^{2}] = 0â‡’ 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,

d

^{2}V/dx^{2}= Ï€/3[-2R – 6x]So, when x = R/3, d

^{2}v/dx^{2}= Ï€/3[-2R – 2R] = -4Ï€R/3 < 0So, 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€r

^{2}hâ‡’ h = 3V/Ï€r

^{2}Slant height of the cone (l) = âˆš(r

^{2 }+ h^{2})l =

And the curved surface area of the cone is

A = Ï€rl

A = Ï€r

A =

On differentiating w.r.t. r, we getdA/dr =

=

For maximum and minimum

Put dA/dr = 0

= 0

â‡’ 2Ï€

^{2}r^{6}– 9V^{2 }= 0â‡’ V

^{2 }= 2Ï€^{2}r^{6}/9â‡’ V

^{ }= âˆš2Ï€^{2}r^{6}/9â‡’ V

^{ }= Ï€r^{3}âˆš2/3or

r = (3V/Ï€âˆš2)

^{1/3}h/r = âˆš2

cotÎ¸ = âˆš2

Semi-vertical angle, Î¸ = cot

^{-1}âˆš2Also, when r < (3V/Ï€âˆš2)

^{1/3}, dA/dr < 0When r > (3V/Ï€âˆš2)

^{1/3}, dA/dr > 0Hence, 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 getdA/dÎ¸ = a

^{2}(2cos2Î¸ + 1/2 Ã— 4cos4Î¸)dA/dÎ¸ = 2a

^{2}(cos2Î¸ + cos4Î¸)

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

^{2}A/dÎ¸^{2}= 2a^{2}(-2sin2Î¸ – 4sin4Î¸)For maximum and minimum

Put dA/dÎ¸ = 0

2a

^{2}(cos2Î¸ + cos4Î¸) = 0cos2Î¸ + cos4Î¸ = 0

cos2Î¸ + 2cos

^{2}2Î¸ – 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, d

^{2}A/dÎ¸^{2}< 0Hence, 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)

The volume of the cylinder that revolve about the breadth,

V = Ï€l

^{2}bV = Ï€(18 – b )

^{2}bV = Ï€(324 + b

^{2 }– 36b)bV = Ï€(324b + b

^{3 }– 36b^{2})

On differentiating w.r.t. b, we getdV/db = Ï€(324 + 3b

^{2 }– 72b)

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

^{2}V/db^{2}= Ï€(6b^{ }– 72)For maximum and minimum

Put dV/db = 0

Ï€(324b + b

^{3 }– 36b^{2}) = 0(b – 6)(b – 18) = 0

b = 6, 18

When b = 6, d

^{2}V/db^{2}= -36Ï€ < 0When b = 18, d

^{2}V/db^{2}= 36Ï€ > 0So, 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(R

^{2 }– x^{2})= âˆš(144 – x

^{2})= (144 – x

^{2})and AD = AO + OD

= R + x = 12 + x

The volume of cone is

V = 1/3 Ï€r

^{2}h= 1/3 Ï€ BD

^{2 }Ã— AD= 1/3 Ï€(144 – x

^{2})(12 + x)= 1/3 Ï€(1728 + 144x – 12x

^{2 }– x^{3})

On differentiating w.r.t. x, we getdV/dx = 1/3Ï€ (144 – 24x – 3x

^{2})For maximum and minimum

Put dV/dx = 0

â‡’ 1/3 Ï€(144 – 24x – 3x

^{2}) = 0â‡’ x = -12, 4

Here x = -12 is not possible

So, x = 4

Now,

d

^{2}V/dx^{2}= Ï€/3(-24 – 6x)At x = 4, d

^{2}v/dx^{2}= -2Ï€(4 + x) = -2Ï€ Ã— 8 = -16Ï€ < 0So, 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 cm^{3}. 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 cm

^{3}Let us assume r and h be the radius and the height of the cylinder.

So, the volume of the cylinder is

V= Ï€r

^{2}h = 2156 …..(i)and the total surface area is

A = 2Ï€rh + 2Ï€r

^{2}A = 2Ï€r (h + r) …..(ii)

So, from eq (i) and (ii)

A = (2156 Ã— 2)/r + 2Ï€r

^{2}

On differentiating w.r.t. r, we getdA/dr = – 4312/4Ï€ + 4Ï€r

For maximum and minimum

Put dA/dr = 0

â‡’ (-4312 + 4Ï€r

^{3})/r^{2}= 0â‡’ r

^{3}= 4312/4Ï€â‡’ r = 7

At r = 7, d

^{2}s/dr^{2}= (8624/r3 + 4Ï€) > 0So, 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€ cm^{3}.

**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(AO

^{2 }– OL^{2})= âˆš(75 – x

^{2})As we know that the volume of cylinder is

V = Ï€r

^{2}h= Ï€AL

^{2 }Ã— ML= Ï€(75 – x

^{2}) Ã— 2x

On differentiating w.r.t. x, we getdV/dx = Ï€[150 – 6x

^{2}]For maximum and minimum

Put dV/dx = 0

â‡’ Ï€[150 – 6x

^{2}] = 0â‡’ x = 5 cm

Also, d

^{2}v/dx^{2}= -12Ï€xAt x = 5, d

^{2}v/dx^{2}= -60Ï€x < 0So, 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€ cm

^{3}

### Question 27. Show that among all positive numbers x and y with x^{2 }+ y^{2 }= r^{2}, the sum x + y is largest when x = y = r/âˆš2.

**Solution:**

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

x

^{2}+y^{2}= r^{2}……(i)Let S be the sum of two positive numbers

S = x + y …..(ii)

= x + âˆš(r

^{2 }– x^{2}) [From eq(ii)]

On differentiating w.r.t. x, we getdS/dx = 1 – x/âˆš(r

^{2 }– x^{2})For maximum and minimum

Put dS/dx = 0

â‡’ 1 – x/âˆš(r

^{2 }– x^{2}) = 0â‡’ x = âˆš(r

^{2}– x^{2})â‡’ 2x

^{2}= r^{2}â‡’ x = r/âˆš2, -r/âˆš2

According to the question x and y are the positive numbers

So, x â‰ -r/âˆš2

Now, d

^{2}S/dx^{2}=At, x = r/âˆš2, d

^{2}S/dx^{2}= < 0So, 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 x^{2} = 4y which are nearest to the point (0, 5).

**Solution:**

The given equation of parabola is

x

^{2}= 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 = x

^{2}+ (y – 5)^{2}…..(ii)Q = 4y + (y – 5)

^{2}

On differentiating w.r.t. y, we getâ‡’ 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

x

^{2}= 12x = 2âˆš3

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

Now,

d

^{2}Q/dy^{2}= 2 > 0So, 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 y^{2 }= 4x which is nearest to the point (2, -8).

**Solution:**

The given equation of the curve is

y

^{2 }= 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)= (y

^{2}/4 – 2)^{2}+ (y + 8)^{2}

On differentiating w.r.t. y, we getdQ/dy = 2(y

^{2}/4 – 2) Ã— y/2 + 2(y + 8)= (y

^{3 }– 8y)/4 + 2y + 16= y

^{3}/4 + 16For maximum and minimum

Put dQ/dy = 0

â‡’ y

^{3}/4 + 16 = 0â‡’ y = -4

Now,

d

^{2}Q/dy^{2}= 3y^{2}/4At y = -4, d

^{2}S/dy^{2}= 12 > 0Ao, y = -4 is the point of local minima

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

x = y

^{2}/4 = 4Hence, the point is(4, -4) which is nearest to (2, -8).

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

**Solution:**

The given equation of the curve is

x

^{2 }= 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}+ (x^{2}/8 – 4)^{2}

On differentiating w.r.t. x, we getdQ/dx = 2(x – 2) + 2(x

^{2}/8 – 4) Ã— 2x/8= 2(x – 2) + (x

^{2}– 32)x/16Also, d

^{2}Q/dx^{2}= 2 + 1/16[x^{2}– 32 + 2x^{2}]= 2 + 1/16[3x

^{2}– 32]For maxima and minima,

dQ/dx = 0

â‡’ 2(x – 2) + x(x

^{2}– 32)/16 = 0â‡’ 32x – 64 + x

^{3}-32x = 0â‡’ x

^{3}– 64 = 0â‡’ x = 4

At x = 4, d

^{2}Q/dx^{2}= 2 + 1/16[16 Ã— 3 – 32] = 2 + 1 = 3 > 0So, x = 4 is point of local minima

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

y = x

^{2}/8 = 2So, the P(4, 2) is the nearest point to (2, 4)

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