NCERT Exemplar: Application of Derivatives- 1 | Mathematics (Maths) Class 12 - JEE PDF Download

Q.1. A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.
Ans.
Ball of salt is spherical
∴ Volume of ball, where r = radius of the ball
As per the question, where S = surface area of the ball
⇒[∵ S = 4pr²]
⇒
⇒(K = Constant of proportionality)
⇒
∴
Hence, the radius of the ball is decreasing at constant rate.

Q.2. If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.
Ans.
We know that:
Area of circle, A = πr², where r = radius of the circle.
and perimeter = 2πr
As per the question,
= K, where K = constant
⇒
∴...(1)
Now Perimeter c = 2πr
Differentiating both sides w.r.t., t, we get
⇒
⇒[From (1)]
⇒
Hence, the perimeter of the circle varies inversely as the radius of the circle.

Q.3. A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is 10 m/s, how fast is the string being let out; when the kite is 250 m away from the boy who is flying the kite? The height of boy is 1.5 m.
Ans.
Given that height of the kite (h) = 151.5 m
Speed of the kite(V) = 10 m/s
Let FD be the height of the kite and AB be the height of the boy.
Let AF = x m
∴ BG = AF = x
m and

From the figure, we get that
GD = DF – GF ⇒ DF - AB
= (151.5 – 1.5) m = 150 m [∵ AB = GF]
Now in ΔBGD,
BG² + GD² = BD² (By Pythagoras Theorem)
⇒ x² + (150)² = (250)²
⇒ x² + 22500 = 62500 ⇒ x² = 62500 – 22500
⇒ x² = 40000 ⇒ x = 200 m
Let initially the length of the string be y m
∴ In ΔBGD
BG² + GD² = BD² ⇒ x² + (150)² = y²
Differentiating both sides w.r.t., t, we get
⇒
⇒ 2 x 200 x 10 = 2 x 250
∴
Hence, the rate of change of the length of the string is 8 m/s.

Q.4. Two  men A and B start with velocities v at the same time from the junction of two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being separated.
Ans.
Let P be any point at which the two roads are inclined at an angle of 45°.
Two men A and B are moving along the roads PA and PB respectively with the same speed ‘V’.
Let A and B be their final positions such that AB = y

∠APB = 45° and they move with the same speed.
∴ ∠APB is an isosceles triangle. Draw  
AB = y ∴and PA = PB = x (let)
∠APQ = ∠BPQ =
[∵ In an isosceles D, the altitude drawn from the vertex, bisects the base]
Now in right ΔAPQ ,

⇒
Differentiating both sides w.r.t, t, we get


Hence, the rate of their separation isunit/s.

Q.5. Find an angle θ, 0 < θ <which increases twice as fast as its sine.
Ans.
As per the given condition,

⇒= 2 cos⇒ 1 = 2 cos θ
∴ cos θ =⇒ cos θ =
Hence, the required angle is

Q.6. Find the approximate value of (1.999)⁵.
Ans.
(1.999)⁵ = (2 – 0.001)⁵ 
Let x = 2 and Δx = – 0.001
Let y = x⁵
Differentiating both sides w.r.t, x, we get
= 5x⁴ = 5(2)⁴= 80
Now Δy == 80 × (- 0.001) = - 0.080

∴ (1.999)⁵ = y + Dy
= x⁵ – 0.080 = (2)⁵ – 0.080 = 32 – 0.080 = 31.92
Hence, approximate value of (1.999)⁵ is 31.92.

Q.7. Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.
Ans.
Internal radius r = 3 cm
and external radius R = r + Δr = 3.0005 cm
∴ Δr = 3.0005 – 3 = 0.0005 cm
Let y = r³ ⇒ y + Δy = (r + Δr)³ = R³ = (3.0005)³ ...(i)
Differentiating both sides w.r.t., r, we get
= 3r²
∴ Δy == 3r² x 0.0005
= 3 x (3)² x 0.0005 = 27 x 0.0005 = 0.0135
∴ (3.0005)³ = y + Δy [From eq. (i)]
= (3)³ + 0.0135 = 27 + 0.0135 = 27.0135
Volume of the shell =

= 4π × 0.005 = 4 x 3.14 x 0.0045 = 0.018 π cm³
Hence, the approximate volume of the metal in the shell is 0.018 π cm³.

Q.8. A man, 2m tall, walks at the rate oftowards a street light which is above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he isfrom the base of the light?
Ans.
Let AB is the height of street light post and CD is the height of the man such that
and CD 2 m

Let  BC = x length (the distance of the man from the lamp post) and CE = y is the length of the shadow of the man at any instant.
From the figure, we see that
ΔABE ~ ΔDCE [by AAA Similarity]
∴ Taking ratio of their corresponding sides, we get

⇒
⇒ 8y = 3x + 3y ⇒ 8y - 3y = 3x ⇒ 5y = 3x
Differentiating both sides w.r.t, t, we get

⇒
[∵ man is moving in opposite direction]
= - 1 m/s
Hence, the length of shadow is decreasing at the rate of 1 m/s.
Now let u = x + y
(u = distance of the tip of shadow from the light post)
Differentiating both sides w.r.t. t, we get

Hence, the tip of the shadow is moving at the rate of
towards the light post and the length of shadow decreasing at the rate of 1 m/s.

Q.9. A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200 (10 – t)². How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?
Ans.
Given that L = 200(10 – t)²
where L represents the number of litres of water in the pool.
Differentiating both sides w.r.t, t, we get∴
= 200 x 2(10 - t) (-1) = - 400(10 - t)
But the rate at which the water is running out
= 400(10 - t)...(1)
Rate at which the water is running after 5 seconds
= 400 x (10 - 5) = 2000 L/s (final rate)
For initial rate put t = 0 = 400(10 - 0) = 4000 L/s
The average rate at which the water is running out

Hence, the required rate = 3000 L/s.

Q.10. The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.
Ans.
Let x  be the length of the cube
∴ Volume of the cube V = x³ ...(1)
Given that
Differentiating Eq. (1) w.r.t. t, we get

∴
Now surface area of the cube, S = 6x² 
Differentiating both sides w.r.t. t, we get

⇒(4K = constant)
Hence, the surface area of the cube varies inversely as the length of the side.

Q.11. x and y are the sides of two squares such that y = x – x² . Find the rate of change of the area of second square with respect to the area of first square.
Ans.
Let area of the first square A₁ = x² 
and area of the second square A₂ = y² 
Now A₁= x² and A₂ = y²  = (x – x²)² 
Differentiating both A₁ and A₂ w.r.t. t, we get

∴

Hence, the rate of change of area of the second square with respect to first is
2x² – 3x + 1.

Q.12. Find the condition that the curves 2x = y² and 2xy = k intersect orthogonally.
Ans.
The two circles intersect orthogonally if the angle between the tangents drawn to the two circles at the point of their intersection is 90°.
Equation of the two circles are given as
2x = y²...(i)
and 2 = k ...(ii)
Differentiating eq. (i) and (ii) w.r.t. x, we get

(m₁ = slope of the tangent)
⇒ 2xy = k

∴
[m₂ = slope of the other tangent]
If the two tangents are perpendicular to each other,
then m₁ x m₂ = – 1
⇒
Now solving 2x = y² [From (i)]
and 2xy = k [From (ii)]
From eq. (ii)
Putting the value of y in eq. (i)

⇒ 8x³ = k² ⇒ 8(1)³ = k² ⇒ 8 = k²
Hence, the required condition is k² = 8.

Q.13. Prove that the curves xy = 4 and x² + y² = 8 touch each other.
Ans.
Given circles are xy = 4 ...(i)
and x² + y² = 8 ...(ii)
Differentiating eq. (i) w.r.t., x

⇒...(iii)
where, m₁ is the slope of the tangent to the curve.
Differentiating eq. (ii) w.r.t. x

where, m₂ is the slope of the tangent to the circle.
To find the point of contact of the two circles

Putting the value of y² in eq. (ii)
x² + x² = 8 ⇒ 2x² = 8 ⇒ x² = 4
∴ x = ±    2
∵ x² = y² ⇒ y = ± 2
∴ The point of contact of the two circles are (2, 2) and ( - 2, 2).

Q.14. Find the co-ordinates of the point on the curve at which tangent is equally inclined to the axes.
Ans.
Equation of curve is given by
Let (x₁, y₁) be the required point on the curve   
∴
Differentiating both sides w.r.t. x₁, we get


Since the tangent to the given curve at (x₁, y₁) is equally inclined to the axes.
∴ Slope of the tangent
So, from eq. (i) we get

Squaring both sides, we get

Putting the value of y₁ in the given equation of the curve.

⇒
Since y₁ = x₁ 
∴ y₁ = 4
Hence, the required point is (4, 4).

Q.15. Find the angle of intersection of the curves y = 4 – x² and y = x².
Ans.
We know that the angle of intersection of two curves is equal to the angle between the tangents drawn to the curves at their point of intersection.
The given curves are y = 4 – x² ... (i)  and    y = x² ...(ii)
Differentiating eq. (i) and (ii) with respect to x, we have

m₁ is the slope of the tangent to the curve (i).
and
m₂ is the slope of the tangent to the curve (ii).
So, m₁ = – 2x and m₂ = 2x
Now solving eq. (i) and (ii) we get
⇒ 4 – x² = x² ⇒ 2x² = 4 ⇒ x² = 2 ⇒ x = ± √2
So, m₁ = - 2 x =- 2√2 and m₂ = 2x = 2√2
Let θ be the  angle of intersection of two curves
∴

∴
Hence, the required angle is tan^-1

Q.16. Prove that the curves y² = 4x and x² + y² – 6x + 1 = 0 touch each other at the point (1, 2).
Ans.
Given that the equation of the two curves are y² = 4x ...(i)
and x² + y² – 6x + 1 = 0 ...(ii)
Differentiating (i) w.r.t. x, we get
Slope of the tangent at (1, 2), = 1
Differentiating (ii) w.r.t. x ⇒ 2x + 2y
⇒
∴ Slope of the tangent at the same point (1, 2)
⇒
We see that m₁ = m₂ = 1 at the point (1, 2).
Hence, the given circles touch each other at the same point (1, 2).

Q.17. Find the equation of the normal lines to the curve 3x² – y² = 8 which are parallel to the line x + 3y = 4.
Ans.
We have equation of the curve 3x² – y² = 8
Differentiating both sides w.r.t. x, we get
⇒
Slope of the tangent to the given curve =
∴ Slope of the normal to the curve =
Now differentiating both sides the given line x + 3y = 4
⇒
Since the normal to the curve is parallel to the given line x + 3y = 4.
∴
Putting the value of y in 3x² – y² = 8, we get
3x² – x² = 8 ⇒ 2x² = 8 ⇒ x² = 4 ⇒ x = ±    2
∴ y = ± 2
∴ The points on the curve are (2, 2) and (– 2, – 2).
Now equation of the normal to the curve at (2, 2) is

⇒ 3y – 6 = – x + 2 ⇒ x + 3y = 8

⇒ 3y + 6 = – x – 2 ⇒ x + 3y = – 8
Hence, the required equations are x + 3y = 8 and x + 3y = – 8 or x + 3y = ± 8.

Q.18. At what points on the curve x² + y² – 2x – 4y + 1 = 0, the tangents are parallel to the y-axis?
Ans.
Given that the equation of the curve is
x² + y² – 2x – 4y + 1 = 0 ...(i)
Differentiating both sides w.r.t. x, we have

⇒
Since the tangent to the curve is parallel to the y-axis.
∴ Slope
So, from eq. (ii) we get

Now putting the value of y in eq. (i), we get
⇒ x² + (2)² – 2x – 8 + 1 = 0
⇒ x² – 2x + 4 – 8 + 1 = 0
⇒ x² – 2x – 3 = 0 ⇒ x² – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0 ⇒ (x - 3) (x + 1) = 0
⇒ x = – 1 or 3
Hence, the required points are (– 1, 2) and (3, 2).

Q.19. Show that the linetouches the curve y = b.

at the point where the curve intersects the axis of y.
Ans.
Given that y = b × e^– x/a, the equation of curve
andthe equation of line.
Let the coordinates of the point where the curve intersects the y-axis be (0, y₁) Now differentiating y = b × e^– x/a both sides w.r.t. x, we get

So, the slope of the tangent, m₁ =
Differentiating both sides w.r.t. x, we get

So, the slope of the line, m₂ =
If the line touches the curve, then m₁ = m₂
⇒
⇒(Taking log on both sides)
⇒⇒ x = 0
Putting x = 0 in equation y = b × e ^– x/a 
⇒ y = b × e⁰ = b
Hence, the given equation of curve intersect at (0, b) i.e. on y-axis.

Q.20. Show that f (x) = 2x + cot^–1x + logis increasing in R.
Ans.
Given that f (x) = 2x + cot^–1x + log
Differentiating both sides w.r.t. x, we get


For increasing function, f ′(x) ≠ 0
⇒
Squaring both sides, we get 4x⁴ + 1 + 4x² ≥ 1 + x²
⇒ 4x⁴ + 4x² - x² ≥ 0 ⇒ 4x⁴ + 3x² ≥ 0 ⇒ x²(4x² + 3) ≠ 0
which is true for any value of x ∈ R.
Hence, the given function is an increasing function over R.

Q.21. Show that for a ≥ 1, f (x) = √3 sinx – cosx – 2ax + b is decreasing in R.
Ans.
Given that: f (x) = √3 sin x - cos x - 2ax + b, a≥ 1
Differentiating both sides w.r.t. x, we get
f ′(x) = √3 cos x + sin x- 2a
For decreasing function, f ′(x) < 0
∴ √3 cos x + sin x- 2a < 0
⇒
⇒

⇒
Since cos x ∈ [ - 1, 1] and a ≥ 1
∴ f ′(x) < 0
Hence, the given function is decreasing in R.

Q.22. Show that f (x) = tan^–1(sinx + cosx) is an increasing function in

Ans.
Given that: f (x) = tan^–1(sin x + cos x) in
Differentiating both sides w.r.t. x, we get  


For an increasing function f  ′(x) ≥ 0
∴
⇒ cos x – sin x ≥ 0
⇒ cos x ≥ sin x, which is true for
Hence, the given function f(x) is an increasing function in

Q.23. At what point, the slope of the curve y = – x³ + 3x² + 9x – 27 is maximum? Also find the maximum slope.
Ans.
Given that: y = – x³ + 3x² + 9x – 27
Differentiating both sides w.r.t. x, we get= – 3x² + 6x + 9
Let slope of the curve
∴ z = -3x² + 6x + 9
Differentiating both sides w.r.t. x, we get= – 6x + 6
For local maxima and local minima,= 0
∴ – 6x + 6 = 0 ⇒ x = 1
⇒
Put x = 1 in equation of the curve y = (– 1)³ + 3(1)² + 9(1) – 27
= – 1 + 3 + 9 – 27 = – 16
Maximum slope = – 3(1)² + 6(1) + 9 = 12
Hence, (1, – 16) is the point at which the
slope of the given curve is maximum and maximum slope = 12.

Q.24. Prove that f (x) = sinx + √3 cosx has maximum value at x =
Ans.
We have: f (x) = sin x + √3 cosx =



(Maxima)
Maximum value of the function at   

Hence, the given function has maximum value atand     the maximum value is 2.

Long Answer (L.A.)
Q.25. If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is
Ans.
Let ΔABC be the right angled triangle in which ∠B = 90°
Let AC = x, BC = y
∴
∠ACB = θ
Let Z  = x + y (given)
Now area of ΔABC,A =


Squaring both sides, we get

Differentiating both sides w.r.t. y we get
...(i)
For local maxima and local minima,


⇒ yZ ≠ 0 ( ∵ y ≠ 0 and Z ≠ 0)
∴ Z - 3y = 0
⇒( ∵ Z = x + y)
⇒ 3y = x + y ⇒ 3y - y = x ⇒ 2y = x
⇒
∴
Differentiating eq. (i) w.r.t. y, we have


Hence, the area of the given triangle is maximum when the angle
between its hypotenuse and a side is

Q.26. Find the points of local maxima, local minima and the points of inflection of the function f (x) = x⁵ – 5x⁴ + 5x³ – 1.  Also find the corresponding local maximum and local minimum values.
Ans.
We have f (x) = x⁵ – 5x⁴ + 5x³ – 1
⇒ f  ′(x) = 5x⁴ - 20x³ + 15x²
For local maxima and local minima, f ′(x) = 0
⇒ 5x⁴ – 20x³ + 15x² = 0 ⇒ 5x²(x² - 4x + 3) = 0
⇒ 5x²(x² - 3x - x + 3) = 0 ⇒ x²(x - 3) (x - 1) = 0
∴ x = 0, x = 1 and x = 3
Now f ″(x) = 20x³ - 60x² + 30x
⇒ f ″(x)_{at x = 0} = 20(0)³ - 60(0)² + 30(0) = 0 which is neither maxima nor minima.
∴ f (x) has the point of inflection at x = 0
f  ″(x)at x = 1 = 20(1)³ - 60(1)² + 30(1)
= 20 - 60 + 30 = - 10 < 0 Maxima
f ″(x)at x = 3 = 20(3)³ - 60(3)² + 30(3)
= 540 - 540 + 90 = 90 > 0 Minima
The maximum value of the function at x = 1
f(x) = (1)⁵ – 5(1)⁴ + 5(1)³ – 1 = 1 – 5 + 5 – 1 = 0
The minimum value at x = 3 is
f (x) = (3)⁵ – 5(3)⁴ + 5(3)³ – 1
= 243 – 405 + 135 – 1 = 378 – 406 = – 28
Hence, the function has its maxima at x = 1 and the maximum value = 0 and it has minimum value at x = 3 and its minimum value is – 28.
x = 0 is the point of inflection.

Q.27. A telephone company in a town has 500 subscribers on its list and collects fixed charges of ₹ 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of ₹ 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?
Ans.
Let us consider that the company increases the annual subscription by ₹ x.
So, x is the number of subscribers who discontinue the services.
∴ Total revenue, R(x) = (500 – x) (300 + x)
= 150000 + 500x – 300x – x²
= – x² + 200x + 150000
Differentiating both sides w.r.t. x, we get R ′(x) = - 2x + 200
For local maxima and local minima, R ′(x) = 0
- 2x + 200 = 0 ⇒ x = 100
R ″(x) =  - 2 < 0 Maxima
So, R(x) is maximum at x = 100
Hence, in order to get maximum profit, the company should increase its annual subscription by ₹ 100.

Q.28. If the straight line x cosα + y sinα = p touches the curve then prove that a² cos²α + b² sin²α = p².
Ans.
The given curve is ...(i)
and the straight line x cos α + y sin α = p ...(ii)
Differentiating eq. (i) w.r.t. x, we get

So the slope of the curve =
Now differentiating eq. (ii) w.r.t. x, we have

∴
So, the slope of the straight line = – cot α
If the line is the tangent to the curve, then

Now from eq. (ii) we have x cos α + y sin α = p

⇒ a² cos² αy + b² sin² αy = b² sin αp
⇒

⇒ a² cos² α + b² sin² α = p × p
Hence, a² cos² α + b² sin² α = p²
Alternate method: 
We know that y = mx + c will touch the ellipse

Here equation of straight line is x cos α + y sin α = p and
that of ellipse is
x cos α + y sin α = p
⇒ y sin α = – x cos α + p
⇒
Comparing with y = mx + c, we get

So, according to the condition, we get c² = a²m² + b²

⇒
Hence, a² cos² α + b² sin² α = p² 
Hence proved.

Q.29. An open box with square base is to be made of a given quantity of card board of area c². Show that the maximum volume of the box iscubic units.
Ans.
Let x  be the length of the side of the square base of the cubical open box and y be its height.
∴ Surface area of the open box
c² = x² + 4xy ...(i)
Now volume of the box, V = x x x x y
⇒ V = x²y


Differentiating both sides w.r.t. x, we get
...(ii)
For local maxima and local minima,

⇒
∴
Now again differentiating eq. (ii) w.r.t. x, we get

Volume of the cubical box (V) = x² y

Hence, the maximum volume of the open box iscubic units.

Q.30. 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. Also find the maximum volume.
Ans.
Let x and y be the length and breadth of a given rectangle ABCD as per question, the rectangle be revolved about side AD which will make a cylinder with radius x and height y.

∴ Volume of the cylinder V = πr²h
 ⇒ V = πx²y ...(i)
Now perimeter of rectangle P = 2(x + y) ⇒ 36 = 2(x + y)
⇒ x + y = 18 ⇒ y = 18 – x ...(ii)
Putting the value of y in eq. (i) we get
V = πx²(18 – x)
⇒ V = π(18x² – x3)
Differentiating both sides w.r.t. x, we get
...(iii)
For local maxima and local minima
∴ p(36x – 3x²) = 0 ⇒ 36x – 3x² = 0
⇒ 3x(12 – x) = 0
⇒ x ≠ 0 ∴ 12 – x = 0 ⇒  x = 12
From eq. (ii) y = 18 – 12 = 6
Differentiating eq. (iii) w.r.t. x, we get
at x = 12 = π(36 – 6 x 12)
= π(36 - 72) = - 36π < 0 maxima
Now volume of the cylinder so formed = πx²y
= π x (12)² x 6  = π x 144 x 6 = 864π cm³
Hence, the required dimensions are 12 cm and 6 cm  and the maximum volume is 864π cm³.

Q.31. If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?
Ans.
Let x  be the edge of the cube and r be the radius of the sphere.
Surface area of cube = 6x²

and surface area of the sphere = 4πr²
∴ 6x² + 4πr² = K(constant) ⇒...(i)
Volume of the cube = x³ and the volume of sphere =
∴ Sum of their volumes (V) = Volume of cube + Volume of sphere
⇒
⇒
Differentiating both sides w.r.t. x, we get


∴...(ii)
For local maxima and local minima,
∴
⇒
x ≠ 0 ∴ x -

Squaring both sides, we get

⇒
∴
Now putting the value of K in eq. (i), we get
6x² + 4πr² = x²(p + 6)
⇒ 6x² + 4πr² = πx² + 6x² ⇒ 4πr² = πx² ⇒ 4r² = x²
∴2r = x
∴ x : 2r = 1 : 1
Now differentiating eq. (ii) w.r.t x, we have

Put

So it is minima.
Hence, the required ratio is 1 : 1 when the combined volume is minimum.

Q.32. AB is a diameter of a circle and C is any point on the circle. Show that the area of ∆ ABC is maximum, when it is isosceles.
Ans.
Let AB be the diameter and C be any point on the circle with radius r.
∠ACB = 90° [angle in the semi circle is 90°]
Let AC = x
∴
⇒...(i)
Now area of
⇒
Squaring both sides, we get

Let A² = Z
∴
Differentiating both sides w.r.t. x, we get
...(ii)
For local maxima and local minima
∴

x ≠ 0 ∴ 2r² - x² = 0 

⇒ x² = 2r² ⇒ x = √2r = AC
Now from eq. (i) we have

So AC = BC
Hence, ΔABC is an isosceles triangle.
Differentiating eq. (ii) w.r.t. x, we get
Put x = √2r
∴

Hence, the area of ΔABC is maximum when it is an isosceles triangle.

Q.33. A metal box with a square base and vertical sides is to contain 1024 cm³. The material for the top and bottom costs Rs 5/cm² and the material for the sides costs Rs 2.50/cm² . Find the least cost of the box.
Ans.
Let x  be the side of the square base and y be the length of the vertical sides.
Area of the base and bottom = 2x² cm² 
∴Cost of the material required = ₹ 5 x 2x² = ₹ 10x²
Area of the 4 sides = 4xy cm²
∴ Cost of the material for the four sides
= ₹ 2.50 x 4xy = ₹ 10xy
Total cost C = 10x² + 10xy ...(i)
New volume of the box = x x x x y
⇒ 1024 = x²y

∴...(ii)
Putting the value of y in eq. (i) we get

Differentiating both sides w.r.t. x, we get
...(iii)
For local maxima and local minima

⇒ 20x³ – 10240 = 0 ⇒ x³ = 512 ⇒ x = 8 cm
Now from eq. (ii)

∴ Cost of material used C = 10x² + 10xy
= 10 x 8 x 8 + 10 x 8 x 16  = 640 + 1280 = 1920
Now differentiating eq. (iii) we get

Put x = 8

Hence, the required cost is ₹1920 which is the minimum.

Q.34. The sum of the surface areas of a rectangular parallelopiped with sides x, 2x andand a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.
Ans.
Let ‘r ’ be the radius of the sphere.
∴ Surface area of the sphere = 4πr²
Volume of the sphere =
The sides of the parallelopiped are x, 2x and
∴ Its surface area =

Volume of the parallelopiped =
As per the conditions of the question,
Surface area of the parallelopiped + Surface area of the sphere = constant
⇒ 6x² + 4πr² = K (constant) ⇒ 4πr² = K - 6x²
∴...(i)
Now let V = Volume of parallelopiped + Volume of the sphere
⇒
⇒[from eq. (i)]
⇒
⇒
⇒
Differentiating both sides w.r.t. x, we have

For local maxima and local minima, we have
∴
⇒
⇒
Here x ≠ 0 and
⇒
Squaring both sides, we get
4πx² = 9(K – 6x²)  ⇒ 4πx² = 9K - 54x²
⇒ 4πx² + 54x² = 9K
⇒ ...(ii)

⇒ 2x²(2π + 27) = 9K
∴
Now from eq. (i) we have

⇒
⇒
⇒
Now we have
Differentiating both sides w.r.t. x, we get

Put 



[ ∵ 27 - 2π > 0]
so, it is minima.

Hence, the sum of volume is minimum for
∴ Minimum volume,


Hence, the required minimum volume isand x = 3r.