# 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

⇒ (l + 2b) + (l + l) = 12

⇒ 3l + 2b = 12 ……(i)

Mow we find the area of the window (A) = Area of the rectangle + Area of the equilateral △

A = l (12 – 3l / 2) + √3/4 l

^{2}

On differentiating w.r.t. l, we getdA/dl = 6 – 3l + (√3/2)l = 6 – √3(√3 – 1/2)l

For maxima and minima,

Put dA/dl = 0

⇒ 6 – √3(√3 – 1/2)l = 0

⇒ l = 6/{√3(√3 – 1/2) } = 12- (6 – √3)

Now, d

^{2}A/dl^{2 }= -√3(√3 – 1/2) = -3 + √3/2So, l = 12/(6 – √3) is the point of local maxima

So, When l = 12/(6 – √3), the area of the window is maximum

From eq(i), we get

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

### Question 17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. 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 √(R

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

V = πr

^{2}h = 2πr^{2}√(R^{2 }– r^{2})

On differentiating w.r.t. r, we getdV/dr = 4πr √(R

^{2 }– r^{2}) += 4πr √(R

^{2 }– r^{2}) –=

=

For maxima and minima,

Put dV/dr = 0

4πrR

^{2}– 6πr^{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 = 2√(R

^{2 }– 2R^{2}/3) = 2√(R^{2}/3) = 2R/√3Hence 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 △OHE,

HE

^{2}= OE^{2}– OH^{2}HE = b = …..(i)

Now we find the area of the EFGH rectangle

A = lb = l ×

A = 1/2 l √(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

⇒

⇒ l = ±√2r

As we know that l can’t be negative so l ≠ -√2r

So, when l = √2r, d

^{2}A/dl^{2 }< 0Hence, the area of the rectangle is maximum when l = √2r

Now put the value of l = √2r in eq(i), we get

Now we find the area of rectangle = lb

= √2r × r/√2

= r

^{2}

### Question 19. Prove that a conical tent of given capacity will require the least amount of canvas when the height is √2 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 πr

^{2}hh = 3V/r

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

A = πrl

Here, l is the slant height = √(r

^{2 }+ h^{2})= πr√(r

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

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

=

For maxima and minima,

Put dA/dr = 0

2π

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

^{6}= 9V^{2}/2π^{2}When, r

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

^{6}= 6V^{2}/2π^{2}Now put r

^{6}= 9V^{2}/2π^{2 }in eq(i), we geth = 3V/πr

^{2}= 3/πr^{2}(2π^{2}r^{6}/9)^{1/2 }= (3/πr^{2})(√2πr^{3}/3) = √2rHence 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 = √(R

^{2 }– x^{2}) and AD = (R + x)Now,

The volume of the cone is

V = 1/3 πr

^{2}h= 1/3 πBD

^{2}× AD= 1/3 π (R

^{2 }– x^{2}) × (R + x)

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

^{2 }– x^{2}]= π/3 [R

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

Put dv/dx = 0

⇒ π/3 [R

^{2}– 2xR – 3x^{2}] = 0⇒ π/3 [(R – 3x) (R + x)] = 0

⇒ R – 3x = 0 or x = -R

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

⇒ x = R/3

Now,

d

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

^{2}v/dx^{2}= π/3[-2R – 2R] = -4πR/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} (√2).

**Solution:**

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

So, the volume of the cone (V) = 1/3 πr

^{2}h⇒ h = 3V/πr

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

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

And the curved surface area of the cone is

A = πrl

A = πr

A =

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

=

For maximum and minimum

Put dA/dr = 0

= 0

⇒ 2π

^{2}r^{6}– 9V^{2 }= 0⇒ V

^{2 }= 2π^{2}r^{6}/9⇒ V

^{ }= √2π^{2}r^{6}/9⇒ V

^{ }= πr^{3}√2/30r

r = (3V/π√2)

^{1/3}h/r = √2

cotθ = √2

Semi-vertical angle, θ = cot

^{-1}√2Also, when r < (3V/π√2)

^{1/3}, dA/dr < 0When r > (3V/π√2)

^{1/3}, dA/dr > 0Hence, the curved surface for r = (3V/π√2)

^{1/3 }is least.

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

**Solution:**

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

and the vertical angle∠BAC = 2θ

Radius of the circle = a

Now, draw AM perpendicular to BC.

From the figure we conclude that in △ABC 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, ∠BOC = 2 × 2θ = 4θ

and ∠COM = 2θ [Since △OMB and △OMC are congruent triangles]

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

In △OMC,

CM = asin2θ and OM = acos2θ

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

BC = 2asin2θ …..(i)

In △ABC,

AM = AO + OM

AM = a + acos2θ …..(ii)

Now the area of △ABC is,

A = 1/2 × BC × AM

= 1/2 × 2asin2θ × (a + acos2θ) ……(iii)

On differentiating w.r.t.θ, we getdA/dθ = a

^{2}(2cos2θ + 1/2 × 4cos4θ)dA/dθ = 2a

^{2}(cos2θ + cos4θ)

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

^{2}A/dθ^{2}= 2a^{2}(-2sin2θ – 4sin4θ)For maximum and minimum

Put dA/dθ = 0

2a

^{2}(cos2θ + cos4θ) = 0cos2θ + cos4θ = 0

cos2θ + 2cos

^{2}2θ – 1 = 0(2cos2θ – 1)(2cos2θ + 1) = 0

cos2θ = 1/2 or cos2θ = -1

2θ = π/3 or 2θ = π

θ = π/6 or θ = π/2

When θ = π/2, it will not form a triangle.

When θ = π/6, d

^{2}A/dθ^{2}< 0Hence, the area of the triangle is maximum when θ = π/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 = πl

^{2}bV = π(18 – b )

^{2}bV = π(324 + b

^{2 }– 36b)bV = π(324b + b

^{3 }– 36b^{2})

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

^{2 }– 72b)

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

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

Put dV/db = 0

π(324b + b

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

b = 6, 18

When b = 6, d

^{2}V/db^{2}= -36π < 0When b = 18, d

^{2}V/db^{2}= 36π > 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 △BOD,

BD = √(R

^{2 }– x^{2})= √(144 – x

^{2})= (144 – x

^{2})and AD = AO + OD

= R + x = 12 + x

The volume of cone is

V = 1/3 πr

^{2}h= 1/3 π BD

^{2 }× AD= 1/3 π(144 – x

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

^{2 }– x^{3})

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

^{2})For maximum and minimum

Put dV/dx = 0

⇒ 1/3 π(144 – 24x – 3x

^{2}) = 0⇒ x = -12, 4

Here x = -12 is not possible

So, x = 4

Now,

d

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

^{2}v/dx^{2}= -2π(4 + x) = -2π × 8 = -16π < 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= πr

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

A = 2πrh + 2πr

^{2}A = 2πr (h + r) …..(ii)

So, from eq (i) and (ii)

A = (2156 × 2)/r + 2πr

^{2}

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

For maximum and minimum

Put dA/dr = 0

⇒ (-4312 + 4πr

^{3})/r^{2}= 0⇒ r

^{3}= 4312/4π⇒ r = 7

At r = 7, d

^{2}s/dr^{2}= (8624/r3 + 4π) > 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 5√3 cm is 500π cm^{3}.

**Solution:**

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

Given that R be the radius of the sphere = 5√3

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

So, h = 2x

Now, In △AOL,

AL = √(AO

^{2 }– OL^{2})= √(75 – x

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

V = πr

^{2}h= πAL

^{2 }× ML= π(75 – x

^{2}) × 2x

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

^{2}]For maximum and minimum

Put dV/dx = 0

⇒ π[150 – 6x

^{2}] = 0⇒ x = 5 cm

Also, d

^{2}v/dx^{2}= -12πxAt x = 5, d

^{2}v/dx^{2}= -60πx < 0So, x = 5 is point of local maxima.

Hence, the volume is maximum when x = 5

So, the maximum volume of cylinder is

= π(75 – 25) × 10 = 500π 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/√2.

**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 + √(r

^{2 }– x^{2}) [From eq(ii)]

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

^{2 }– x^{2})For maximum and minimum

Put dS/dx = 0

⇒ 1 – x/√(r

^{2 }– x^{2}) = 0⇒ x = √(r

^{2}– x^{2})⇒ 2x

^{2}= r^{2}⇒ x = r/√2, -r/√2

According to the question x and y are the positive numbers

So, x ≠ -r/√2

Now, d

^{2}S/dx^{2}=At, x = r/√2, d

^{2}S/dx^{2}= < 0So, x = r/√2 is point of local maxima.

Hence, the sum is largest when x = y = r/√2

### 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 get⇒ dQ/dy = 4 + 2(y – 5)

For maximum and minimum

Put dQ/dy = 0

⇒ 4 + 2(y – 5) = 0

⇒ 2y = 6

⇒ y = 3

From eq(i), we get

x

^{2}= 12x = 2√3

⇒ P = (2√3, 3) and p’ = (-2√3, 3)

Now,

d

^{2}Q/dy^{2}= 2 > 0So, P and P’ are the point of local minima.

Hence, the nearest points are P(2√3, 3) and P”(2√3, 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) × y/2 + 2(y + 8)= (y

^{3 }– 8y)/4 + 2y + 16= y

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

Put dQ/dy = 0

⇒ y

^{3}/4 + 16 = 0⇒ 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) × 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

⇒ 2(x – 2) + x(x

^{2}– 32)/16 = 0⇒ 32x – 64 + x

^{3}-32x = 0⇒ x

^{3}– 64 = 0⇒ x = 4

At x = 4, d

^{2}Q/dx^{2}= 2 + 1/16[16 × 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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