Class 12 RD Sharma Solutions β Chapter 18 Maxima and Minima β 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
β (l + 2b) + (l + l) = 12
β 3l + 2b = 12 β¦β¦(i)
Mow we find the area of the window (A) = Area of the rectangle + Area of the equilateral β³
A = l (12 β 3l / 2) + β3/4 l2
On differentiating w.r.t. l, we get
dA/dl = 6 β 3l + (β3/2)l = 6 β β3(β3 β 1/2)l
For maxima and minima,
Put dA/dl = 0
β 6 β β3(β3 β 1/2)l = 0
β l = 6/{β3(β3 β 1/2) } = 12- (6 β β3)
Now, d2A/dl2 = -β3(β3 β 1/2) = -3 + β3/2
So, l = 12/(6 β β3) is the point of local maxima
So, When l = 12/(6 β β3), the area of the window is maximum
From eq(i), we get
b = (12 β 3l)/2 = [12 β 3{12/(6 β β3)}]/2 = (24 β 6β3)/(6 β β3)
Question 17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/β3. 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 β(R2 β r2)
Now we find the volume of the cylinder is
V = Οr2h = 2Οr2β(R2 β r2)
On differentiating w.r.t. r, we get
dV/dr = 4Οr β(R2 β r2) +
= 4Οr β(R2 β r2) β
=
=
For maxima and minima,
Put dV/dr = 0
4ΟrR2 β 6Οr3 = 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 = 2β(R2 β 2R2/3) = 2β(R2/3) = 2R/β3
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 β³OHE,
HE2 = OE2 β OH2
HE = b = β¦..(i)
Now we find the area of the EFGH rectangle
A = lb = l Γ
A = 1/2 l β(4r2 β 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
β
β l = Β±β2r
As we know that l canβt be negative so l β -β2r
So, when l = β2r, d2A/dl2 < 0
Hence, the area of the rectangle is maximum when l = β2r
Now put the value of l = β2r in eq(i), we get
Now we find the area of rectangle = lb
= β2r Γ r/β2
= r2
Question 19. Prove that a conical tent of given capacity will require the least amount of canvas when the height is β2 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 Οr2h
h = 3V/r2 β¦β¦(i)
And the surface area of the cone is
A = Οrl
Here, l is the slant height = β(r2 + h2)
= Οrβ(r2 + h2)
=
On differentiating w.r.t. r, we get
dA/dr =
=
For maxima and minima,
Put dA/dr = 0
2Ο2r6 = 9V2
r6 = 9V2/2Ο2
When, r6 = 9V2/2Ο2, d2S/dr2 > 0
Hence, the surface area of the cone is the least when r6 = 6V2/2Ο2
Now put r6 = 9V2/2Ο2 in eq(i), we get
h = 3V/Οr2 = 3/Οr2(2Ο2r6/9)1/2 = (3/Οr2)(β2Οr3/3) = β2r
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 = β(R2 β x2) and AD = (R + x)
Now,
The volume of the cone is
V = 1/3 Οr2h
= 1/3 ΟBD2 Γ AD
= 1/3 Ο (R2 β x2) Γ (R + x)
On differentiating w.r.t. x, we get
dV/dx = Ο/3 [-2x (R + x) + R2 β x2]
= Ο/3 [R2 β 2xR β 3x2]
For maximum and minimum
Put dv/dx = 0
β Ο/3 [R2β 2xR β 3x2] = 0
β Ο/3 [(R β 3x) (R + x)] = 0
β R β 3x = 0 or x = -R
Here, x = -R is not possible because -r will make the altitude 0
β x = R/3
Now,
d2V/dx2 = Ο/3[-2R β 6x]
So, when x = R/3, d2v/dx2 = Ο/3[-2R β 2R] = -4ΟR/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 (β2).
Solution:
Let us assume h, r and ΞΈ be the height, radius and semi vertical angle of the right-angled triangle.
So, the volume of the cone (V) = 1/3 Οr2h
β h = 3V/Οr2
Slant height of the cone (l) = β(r2 + h2)
l =
And the curved surface area of the cone is
A = Οrl
A = Οr
A =
On differentiating w.r.t. r, we get
dA/dr =
=
For maximum and minimum
Put dA/dr = 0
= 0
β 2Ο2r6 β 9V2 = 0
β V2 = 2Ο2r6/9
β V = β2Ο2r6/9
β V = Οr3β2/3
or
r = (3V/Οβ2)1/3
h/r = β2
cotΞΈ = β2
Semi-vertical angle, ΞΈ = cot-1β2
Also, when r < (3V/Οβ2)1/3, dA/dr < 0
When r > (3V/Οβ2)1/3, dA/dr > 0
Hence, the curved surface for r = (3V/Οβ2)1/3 is least.
Question 22 An isosceles triangle of vertical angle 2ΞΈ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when ΞΈ = Ο/6.
Solution:
Let us considered ABC is an isosceles triangle such that AB = AC
and the vertical angleβ BAC = 2ΞΈ
Radius of the circle = a
Now, draw AM perpendicular to BC.
From the figure we conclude that in β³ABC 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, β BOC = 2 Γ 2ΞΈ = 4ΞΈ
and β COM = 2ΞΈ [Since β³OMB and β³OMC are congruent triangles]
OA = OB = OC =a [Radius of the circle]
In β³OMC,
CM = asin2ΞΈ and OM = acos2ΞΈ
BC = 2CM [Perpendicular from the centre bisects the chord]
BC = 2asin2ΞΈ β¦..(i)
In β³ABC,
AM = AO + OM
AM = a + acos2ΞΈ β¦..(ii)
Now the area of β³ABC is,
A = 1/2 Γ BC Γ AM
= 1/2 Γ 2asin2ΞΈ Γ (a + acos2ΞΈ) β¦β¦(iii)
On differentiating w.r.t. ΞΈ, we get
dA/dΞΈ = a2(2cos2ΞΈ + 1/2 Γ 4cos4ΞΈ)
dA/dΞΈ = 2a2 (cos2ΞΈ + cos4ΞΈ)
Again differentiating w.r.t. ΞΈ, we get
d2A/dΞΈ2 = 2a2(-2sin2ΞΈ β 4sin4ΞΈ)
For maximum and minimum
Put dA/dΞΈ = 0
2a2(cos2ΞΈ + cos4ΞΈ) = 0
cos2ΞΈ + cos4ΞΈ = 0
cos2ΞΈ + 2cos22ΞΈ β 1 = 0
(2cos2ΞΈ β 1)(2cos2ΞΈ + 1) = 0
cos2ΞΈ = 1/2 or cos2ΞΈ = -1
2ΞΈ = Ο/3 or 2ΞΈ = Ο
ΞΈ = Ο/6 or ΞΈ = Ο/2
When ΞΈ = Ο/2, it will not form a triangle.
When ΞΈ = Ο/6, d2A/dΞΈ2 < 0
Hence, the area of the triangle is maximum when ΞΈ = Ο/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 β b β¦β¦(i)
The volume of the cylinder that revolve about the breadth,
V = Οl2b
V = Ο(18 β b )2b
V = Ο(324 + b2 β 36b)b
V = Ο(324b + b3 β 36b2)
On differentiating w.r.t. b, we get
dV/db = Ο(324 + 3b2 β 72b)
Again differentiating w.r.t. b, we get
d2V/db2 = Ο(6b β 72)
For maximum and minimum
Put dV/db = 0
Ο(324b + b3 β 36b2) = 0
(b β 6)(b β 18) = 0
b = 6, 18
When b = 6, d2V/db2 = -36Ο < 0
When b = 18, d2V/db2 = 36Ο > 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 β 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 β³BOD,
BD = β(R2 β x2)
= β(144 β x2)
= (144 β x2)
and AD = AO + OD
= R + x = 12 + x
The volume of cone is
V = 1/3 Οr2h
= 1/3 Ο BD2 Γ AD
= 1/3 Ο(144 β x2)(12 + x)
= 1/3 Ο(1728 + 144x β 12x2 β x3)
On differentiating w.r.t. x, we get
dV/dx = 1/3Ο (144 β 24x β 3x2)
For maximum and minimum
Put dV/dx = 0
β 1/3 Ο(144 β 24x β 3x2) = 0
β x = -12, 4
Here x = -12 is not possible
So, x = 4
Now,
d2V/dx2 = Ο/3(-24 β 6x)
At x = 4, d2v/dx2 = -2Ο(4 + x) = -2Ο Γ 8 = -16Ο < 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= Οr2h = 2156 β¦..(i)
and the total surface area is
A = 2Οrh + 2Οr2
A = 2Οr (h + r) β¦..(ii)
So, from eq (i) and (ii)
A = (2156 Γ 2)/r + 2Οr2
On differentiating w.r.t. r, we get
dA/dr = β 4312/4Ο + 4Οr
For maximum and minimum
Put dA/dr = 0
β (-4312 + 4Οr3)/r2 = 0
β r3 = 4312/4Ο
β r = 7
At r = 7, d2s/dr2 = (8624/r3 + 4Ο) > 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 5β3 cm is 500Ο cm3.
Solution:
Let r and h be the radius and height of the cylinder.
Given that R be the radius of the sphere = 5β3
So, from the figure LM = h, OL = x
So, h = 2x
Now, In β³AOL,
AL = β(AO2 β OL2)
= β(75 β x2)
As we know that the volume of cylinder is
V = Οr2h
= ΟAL2 Γ ML
= Ο(75 β x2) Γ 2x
On differentiating w.r.t. x, we get
dV/dx = Ο[150 β 6x2]
For maximum and minimum
Put dV/dx = 0
β Ο[150 β 6x2] = 0
β x = 5 cm
Also, d2v/dx2 = -12Οx
At x = 5, d2v/dx2 = -60Οx < 0
So, x = 5 is point of local maxima.
Hence, the volume is maximum when x = 5
So, the maximum volume of cylinder is
= Ο(75 β 25) Γ 10 = 500Ο 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/β2.
Solution:
Let us considered the two positive numbers are x and y with
x2+y2 = r2 β¦β¦(i)
Let S be the sum of two positive numbers
S = x + y β¦..(ii)
= x + β(r2 β x2) [From eq(ii)]
On differentiating w.r.t. x, we get
dS/dx = 1 β x/β(r2 β x2)
For maximum and minimum
Put dS/dx = 0
β 1 β x/β(r2 β x2) = 0
β x = β(r2 β x2)
β 2x2 = r2
β x = r/β2, -r/β2
According to the question x and y are the positive numbers
So, x β -r/β2
Now, d2S/dx2 =
At, x = r/β2, d2S/dx2= < 0
So, x = r/β2 is point of local maxima.
Hence, the sum is largest when x = y = r/β2
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 β¦β¦.(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 β 5)2 β¦..(ii)
Q = 4y + (y β 5)2
On differentiating w.r.t. y, we get
β dQ/dy = 4 + 2(y β 5)
For maximum and minimum
Put dQ/dy = 0
β 4 + 2(y β 5) = 0
β 2y = 6
β y = 3
From eq(i), we get
x2 = 12
x = 2β3
β P = (2β3, 3) and pβ = (-2β3, 3)
Now,
d2Q/dy2 = 2 > 0
So, P and Pβ are the point of local minima.
Hence, the nearest points are P(2β3, 3) and Pβ(2β3, 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 β¦.(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 β 2)2 + (y + 8)2 β¦β¦.(ii)
= (y2/4 β 2)2 + (y + 8)2
On differentiating w.r.t. y, we get
dQ/dy = 2(y2/4 β 2) Γ y/2 + 2(y + 8)
= (y3 β 8y)/4 + 2y + 16
= y3/4 + 16
For maximum and minimum
Put dQ/dy = 0
β y3/4 + 16 = 0
β 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 β¦.(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 β 2)2 + (y β 4)2 β¦β¦..(ii)
= (x β 2)2 + (x2/8 β 4)2
On differentiating w.r.t. x, we get
dQ/dx = 2(x β 2) + 2(x2/8 β 4) Γ 2x/8
= 2(x β 2) + (x2 β 32)x/16
Also, d2Q/dx2 = 2 + 1/16[x2 β 32 + 2x2]
= 2 + 1/16[3x2 β 32]
For maxima and minima,
dQ/dx = 0
β 2(x β 2) + x(x2 β 32)/16 = 0
β 32x β 64 + x3 -32x = 0
β x3 β 64 = 0
β x = 4
At x = 4, d2Q/dx2 = 2 + 1/16[16 Γ 3 β 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)