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## Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 6 Application of Derivatives

### Plus Two Maths Application of Derivatives Three Mark Questions and Answers

Question 1.

Find the equation of tangents and normals to the given curves x = cost, y = sin t at t = \(\frac{π}{4}\).

Answer:

Given; x = cost, y = sin t

Equation of tangent at t = \(\frac{π}{4}\) is;

Equation of normal at t = \(\frac{π}{4}\) is;

⇒ \(\sqrt{2}\)y + \(\sqrt{2}\)x = 0 ⇒ y + x = 0.

Question 2.

A ladder Sm long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the decreasing when the foot of the ladder is 4m away from the wall?

Answer:

From the figure we have;

x^{2} + y^{2} = 25 ____(1)

Differentiating w.r.t t;

From (1) when x = 4 ⇒ 16 + y^{2} = 25 ⇒ y = 3

Given; \(\frac{d x}{d t}\) = 2cm/s = 0.02 m/s

(2) ⇒ 4(0.02) + 3 \(\frac{d x}{d t}\) = 0

Question 3.

Find the points on the curve y = x^{3}, the tangents at which are inclined at an angle of 60° to x-axis?

Answer:

\(\frac{d y}{d x}\) = 3x^{2}

Slope of the tangent = tan60°

i.e. 3x^{2} = \(\sqrt{3}\)

Question 4.

Find the equation of the tangent to the parabola y^{2} = 4x + 5 which is parallel to y = 2x + 7.

Answer:

y^{2} = 4x + 5 _____(1)

2y \(\frac{d y}{d x}\) = 4

\(\frac{d y}{d x}\) = \(\frac{4}{2y}\) = \(\frac{2}{y}\)

Given tangent is parallel to y = 2x + 7

ie. Slope of the tangent is 2 ⇒ \(\frac{2}{y}\) = 2 ⇒ y = 1

∴ from (1) ⇒ 1 = 4x+ 5 ⇒ 4x = -4 ⇒ x = -1

So the point of contact is (-1, 1).

∴ Equation of tangent is

y -1 = 2(x + 1) ⇒ y = 2x + 3.

Question 5.

Find the intervals in which the function f given f(x) = 2x^{2} – 3x is

- Strictly increasing.
- Strictly decreasing.

Answer:

Given; f(x) = 2x^{2} – 3x ⇒ f'(x) = 4x – 3

For turning points; f'(x) = 0

⇒ 4x – 3 = 0 ⇒ x = \(\frac{3}{4}\)

The intervals are \(\left(-\infty, \frac{3}{4}\right),\left(\frac{3}{4}, \infty\right)\)

f'(0) = – 3 < 0

∴ Strictly decreasing in \(\left(-\infty, \frac{3}{4}\right)\)

f'(1) = 1 > 0

∴ Strictly increasing in \(\left(\frac{3}{4}, \infty\right)\).

Question 6.

Find the intervals in which the function f(x) = (x + 1)^{3} (x – 3)^{3} strictly increasing or decreasing.

Answer:

Given; f(x) = (x + 1)^{3} (x – 3)^{3}

⇒ f'(x) = (x + 1)^{3} 3(x – 3)^{2} + (x – 3)^{3}3(x + 1)^{2}

= 3(x + 1)^{2}(x – 3)^{2}(x + 1 + x – 3)

= 3(x + 1)^{2}(x – 3)^{2}(2x – 2)

= 6(x +1)^{2} (x – 3)^{2} (x -1)

⇒ 6(x +1)^{2} (x – 3)^{2} (x – 1) = 0

⇒ x = -1, 1, 3

The intervals are

(-∞, -1), (-1, 1), (1, 3), (3, ∞)

f'(-2) = (-2 – 1) < 0

∴ Strictly decreasing in (-∞, -1)

f'(0) = (0 – 1) < 0

∴ Strictly decreasing in (-1, 1)

f'(2) = (2 – 1) > 0

∴ Strictly increasing in (1, 3)

f'(4) = (4 – 1) > 0

∴ Strictly increasing in (3, ∞).

Question 7.

Find the intervals in which the function f(x) = x + \(\frac{1}{x}\) strictly increasing or decreasing.

Answer:

⇒ x = ±1

The intervals are (-∞, -1), (-1, 1), (1, ∞)

f'(-2) > 0

∴ Strictly increasing in (-∞, -1)

f'(0) < 0

∴ Strictly decreasing in (-1, 1)

f'(2) > 0

∴ Strictly increasing in (1, ∞).

Question 8.

Determine whether the f(x) = x^{2} function is strictly monotonic on the indicated interval.

- (-1, 1)
- (-1, 0)
- (0, 1)

Answer:

f(x) = x^{2}

⇒ f'(x) = 2x

⇒ f'(x) = 0 ⇒ 2x = 0 ⇒ x = 0

This turning point divides the domain into the intervals (-∞, 0); (0, ∞).

- Interval (-1,1) f'(x) < 0 and f'(x) > 0. So f(x) is not monotonic.
- Interval (-1,0), f'(x) < 0. ∴ f(x) is strictly monotonic.
- Interval (0, 1) f'(x) > 0 and f(x) is strictly monotonic.

Question 9.

Determine whether the f(x) = x^{3} – x function is strictly monotonic on the indicated interval.

- (-1, 0)
- (-1, -1/2)
- (-1, 1)

Answer:

(x) = x^{3} -x ⇒ f'(x) = 3x^{2} – 1

⇒ f'(x) = 0 ⇒ 3x^{2} – 1 = 0 ⇒ x = ±\(\frac{1}{\sqrt{3}}\)

This turning point divides the domain into the intervals (-∞, \(\frac{1}{\sqrt{3}}\)); (-\(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\)); (\(\frac{1}{\sqrt{3}}\), ∞).

- Interval (-1, 0), f'(x) changes sign. So not monotonic.
- Interval (-1, -1/2), f'(x) > 0 strictly monotonic.
- lnterval(-1, 1) not monotonic

Question 10.

Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 1%.

Answer:

We have; V = x^{3} and ∆x = 1% of x= 0.01x

dV = \(\frac{d V}{d x}\) ∆x = 3x^{2}∆x

= 3x^{2} × 0.01x = 0.03x^{3} = 0.03V

⇒ \(\frac{d V}{V}\) = 0.03

Therefore 3% is the approximate increase in volume.

Question 11.

If the radius of a sphere is measured as 7m with an error of 0.02m then find the approximate error in calculating its volume.

Answer:

Let r be the radius of the sphere and ∆r be the error in measuring the radius then r =7m and ∆r = 0.02 m

We have; V = \(\frac{4}{3}\) πr^{3}

dV = \(\frac{d V}{dr}\) ∆r = \(\frac{4}{3}\) π3r^{2} × ∆r

= 4π(7)^{2} × 0.02 = 3.92 π m^{3}.

Question 12.

The length of a rectangle is decreasing at the rate of 5 cm/min and the width is increasing at the rate of 4cm/min. When length is 8 cm and width is 6 cm, find the rate of change of its area.

Answer:

Let length = x and width = y

Question 13.

Find the equation of tangents and normals to the given curves y= x^{4} – 6x^{3} + 13x^{2} – 10x + 5 at (0, 5)

Answer:

Given; y = x^{4} – 6x^{3} + 13x^{2} – 10x + 5 at (0, 5)

⇒ \(\frac{d y}{d x}\) = 4x^{3} – 18x^{2} + 26x – 10

Slope = \(\left(\frac{d y}{d x}\right)_{x=0}\) = -10

Equation of tangent at (0, 5) is;

y – 5 =(-10)(x – 0)

⇒ y – 5 = -10x ⇒ 10x + y – 5 = 0

Equation of normal at (0, 5) is;

y – 5 = \(\frac{1}{10}\)(x – 0)

⇒ 10y – 50 = x ⇒ x – 10y + 50 = 0.

Question 14.

Find the equation of tangents and normals to the given curves y = x^{3} at (1, 1)

Answer:

Given; y = x^{3}

⇒ \(\frac{d y}{d x}\) = 3x^{2}

Slope = \(\left(\frac{d y}{d x}\right)_{x=1}\) = 3

Equation of tangent at (1, 1) is; y -1 = (3)(x – 1)

⇒ y – 1 = 3x – 3 ⇒ 3x – y – 2 = 0

Equation of normal at (1, 1) is;

y – 1 = \(-\frac{1}{3}\)(x – 1)

⇒ 3y – 3 = -x + 1 ⇒ x+ 3y – 4 = 0.

Question 15.

The volume of a cube is increasing at the rate of 8cm^{3}/s. How fast is the surface area increasing when the length of an edge is 12cm.

Answer:

Let V be the volume of the cube of side x.

We have volume = V = x^{3}

Rate of change of volume with respect to time ‘t’ is;

ie; differentiating w.r.t t; \(\frac{d V}{d t}\) = 3x^{2}\(\frac{d x}{d t}\)

Given; \(\frac{d V}{d t}\) = 8 and x = 12 ⇒ 8 = 3(12)^{2}\(\frac{d x}{d t}\)

Now let Surface area = S = 6x^{2}

Differentiating w.r.t t;

Question 16.

Find the intervals in which the function f(x) = -2x^{3} – 9x^{2} – 12x + 1 strictly increasing or decreasing.

Answer:

Given; f(x) = -2x^{3} – 9x^{2} – 12x + 1

⇒ f'(x) = -6x^{2} – 18x – 12

= – 6(x^{2} + 3x + 2)

= – 6(x + 2)(x +1)

⇒ f'(x) = 0 ⇒ -6(x + 2)(x +1) = 0

⇒ x = -2, -1

The intervals are (-∞, -2),(- 2, -1),(-1, ∞)

f'(-3) = -(-3 + 2)(-3 + 1) < 0

∴ Strictly decreasing in (-∞, -2)

f'(-1.5) = -(-1.5 + 2)(-1.5 + 1) > 0

∴ Strictly increasing in (- 2, -1)

f'(0) = -(0 + 2)(0 + 1) < 0

Strictly decreasing 1n(-1, ∞).

Question 17.

Find the local maxima and minima of the following functions. Also find the local maximum and minimum values. (each question carry 3 score)

- f(x) = sin x + cosx, 0 < x < \(\frac{\pi}{2}\)
- f(x) = x
^{3}– 3x - f(x) = x
^{3}– 6x^{2}+ 9x + 15 - g(x) = \(\frac{x}{2}\) + \(\frac{2}{x}\), x > 0
- g(x) = \(\frac{1}{x^{2}+2}\)

Answer:

1. Given; f(x) = sinx + cosx

⇒ f'(x) = cosx – sinx

For turning point f'(x) = 0

⇒ cosx – sinx = 0

⇒ cosx = sinx

⇒ x = \(\frac{\pi}{4}\)

f”(x) = -sin x – cosx

⇒ f (\(\frac{\pi}{4}\)) = -sin\(\frac{\pi}{4}\) – cos\(\frac{\pi}{4}\) < 0

Hence f(x) has a local maximum at x = \(\frac{\pi}{4}\) and local maximum value is

2. Given; f(x) = x^{3} – 3x

⇒ f'(x) = 3x^{2} – 3

For turning point f'(x) = 0

⇒ 3x^{2} – 3 = 0

⇒ x = ±1

f”(x) = 6x

When x = -1

⇒ f”(-1) = -6 < 0

Hence f(x) has a local maximum at x = -1 and local maximum value is

f(-1) = (-1)^{3} – 3(-1) = -1 + 3 = 2

When x = 1

⇒ f”(1) = 6 > 0

Hence f(x) has a local minimum at x = 1 and local minimum value is

f(1) = (1)^{3} – 3(1) = 1 – 3 = -2.

3. Given; f(x) = x^{3} – 6x^{2} + 9x + 15

⇒ f'(x) = 3x^{2} – 12x + 9

For turning point f'(x) = 0

⇒ 3x^{2} – 12x + 9 = 0 ⇒ 3(x^{2} – 4x + 3) = 0

⇒ 3(x – 1)(x – 3) = 0 ⇒ x = 1, 3

f”(x) = 6x – 12

When x = 1

⇒ f”( 1) = 6 – 12 < 0

Hence f(x) has a local maximum at x = 1 and local maximum value is

f(1) = (1)^{3} – 6(1)^{2} + 9(1) + 15 = 19

When x = 3

⇒ f”(3) = 6(3) – 12 > 0

Hence f(x) has a local minimum at x = 3 and local minimum value is

f(3) = (3)^{3} – 6(3)^{2} + 9(3) + 15 = 15.

4. Given; g(x) = \(\frac{x}{2}\) + \(\frac{2}{x}\)

⇒ g'(x) = \(\frac{1}{2}\) – \(\frac{2}{x^{2}}\)

For turning point g'(x) = 0

Since x > 0, the acceptable value of x = 2

Hence g(x) has a local maximum at x = 2 and local maximum value is g(2) = \(\frac{2}{2}\) + \(\frac{2}{2}\) = 2

5. Given; g(x) = \(\frac{1}{x^{2}+2}\)

For turning point g'(x) = 0

Hence g(x) has a local maximum at x = 2 and maximum value is g(2) = \(\frac{1}{0+2}=\frac{1}{2}\).

Question 18.

Find the absolute maximum value and minimum value of the following functions.

- f(x) = x
^{3}, x ∈ [-2, 2] - f(x) = 4x – \(\frac{x^{2}}{2}\), x ∈ \(\left[-2, \frac{9}{2}\right]\)

Answer:

1. Given; f(x) = x^{3} ⇒ f'(x) = 3x^{2}

For turning point f'(x) = 0 ⇒ 3x^{2} = 0 ⇒ x = 0

f(- 2) = (-2 )^{3} = -8

f( 2) = (2)^{3} = 8

f(0) = (0)^{3} = 0

Absolute maximum = max{-8, 8, 0} = 8

Absolute minimum = min {-8, 8, 0} = – 8

2. Given; f(x) = 4x – \(\frac{x^{2}}{2}\) ⇒ f'(x) = 4 – x

For turning point f'(x) = 0 ⇒ 4 – x = 0

⇒ x = 4

Absolute maximum = max{-10, 8, 7.875} = 8

Absolute minimum = min {-10, 8, 7.87} = -10.

Question 19.

A television camera at ground level is filming the lift-off of a space shuttle that is rising vertically according to position equation S = 50t^{2}.

The camera is 2000 feet from the launch pad. Find the rate of change in the angle of elevation of the camera 10 seconds after lift-off.

Answer:

Question 20.

Determine whether the f(x) = sinx function is strictly monotonic on the indicated interval.

- (0, 2π)
- (0, π)
- (-π/2, π/2)

Answer:

f(x) = sinx ⇒ f'(x) = cosx changes sign.

- Interval (0, 2π). ∴ f(x) is not monotonic.
- f'(x) changes sign in (0, π) not monotonic.
- f'(x) > 0 in (-π/2, π/2), ie. f(x) is strictly monotonic.

Question 21.

Find the approximate change in the Surface Area of a cube of side x meters caused by decreasing the side by 1%.

Answer:

We have;

S = 6x^{2} and ∆x = 1% of x = -0.01x

dS = \(\frac{d S}{d x}\) ∆x = 6 × 2x × ∆x

= 6 × 2x × -0.01x = -0.02 × 6x^{2} = -0.02S

⇒ \(\frac{d S}{S}\) = -0.02

Therefore 2% is the approximate decrease in surface area.

### Plus Two Maths Application of Derivatives Four Mark Questions and Answers

Question 1.

The length ‘x’ of a rectangle is decreasing at the rate of 2 cm/s and the width ‘y’ is increasing at the rate of 2 cm/s.

- Find the rate of change of Perimeter.
- Find \(\frac{d A}{dt}\) when x = 12 cm and y = 5 cm.

Answer:

Since the length ‘x’ is decreasing and the width ‘y’ is increasing, we have \(\frac{d x}{dt}\) = -2 cm/s and \(\frac{d y}{dt}\)

= 2 cm/sec.

1. The Perimeter ‘P’ of the rectangle is given by

P = 2 (x + y)

2. The area ‘A’ of the rectangle ‘A’ is given by

A = x.y

= 12(2) + 5(-2)

= 24 – 10 = 14 cm2/s.

Question 2.

Find the equation of all lines having slope -1. Which are tangents to the curve?

y = \(\frac{1}{x-1}\), x ≠ 1

Answer:

⇒ x^{2} – 2x = 0 ⇒ x(x – 2) = 0 ⇒ x = 0, x = 2

At x = 0, y = -1

Equation of tangent at (0, -1) is;

At x = 2, y = \(\frac{1}{2-1}\) = 1

Equation of tangent at (2, 1) is; y – 1 = -1(x – 2)

⇒ y – 1 = -x + 2 ⇒ x + y – 3 = 0.

Question 3.

Find the points on the curve x^{2} + y^{2} – 2x – 3 = 0 at which the tangent are parallel to x-axis.

Answer:

Given; x^{2} + y^{2} – 2x – 3 = 0

Differentiating with respect to x;

Since the tangent is parallel to x-axis \(\frac{d y}{d x}\) = 0

\(\frac{1-x}{y}\) = 0 ⇒ x = 1

We have; (1)^{2} + y^{2} – 2(1) – 3 = 0

⇒ y^{2} = 4 ⇒ y = ±2

Hence the points are (1, 2), (1, -2).

Question 4.

Find the equation of the tangent to the curve y = \(\sqrt{3 x-2}\) which is parallel to the line 4x – 2y + 5 = 0.

Answer:

Slope of the line 4x – 2y + 5 = 0 is 2.

acceptable since y is positive.

Hence the point is \(\left(\frac{41}{48}, \frac{3}{4}\right)\)

Equation of tangent is;

⇒ 6(4y – 3) = (48x – 41)

⇒ 24y – 18 = 48x – 41

⇒ 48x – 24y – 23 = 0.

Question 5.

Prove that the curve x = y^{2} and xy = k cut at right angles, if 8k^{2} = 1.

Answer:

x = y^{2} ___(1)

⇒ 1 = 2y \(\frac{d y}{d x}\) ⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2y}\)

xy = 2k ___(2)

⇒ x \(\frac{d y}{d x}\) + y.1 = 0 ⇒ \(\frac{d y}{d x}\) = \(-\frac{y}{x}\)

The product of the slopes will be – 1.

Question 6.

The gradient at any point (x, y) of a curve is 3x^{2} – 12 and the curve through the point (2, -7).

- Find the equation of the tangent at the point ( 2, -7 ). (2)
- Find the equation to the curve. (2)

Answer:

1. Given gradient as 3x^{2} – 12 ⇒ \(\frac{d y}{d x}\) = 3x^{2} – 12

Slope at (2, -7) is given by

\(\left(\frac{d y}{d x}\right)_{x=2}\) = 3(2)^{2} – 12 = 0

Since slope is zero, the tangent is parallel to x – axis.

Here y = – 7 is the equation of the tangent at (2, -7).

2. Given, \(\frac{d y}{d x}\) = 3x^{2} – 12

⇒ ∫dy = ∫(3x^{2} – 12 )dx

y = 3\(\frac{x^{3}}{3}\) – 12x + c ⇒ y = x^{3} – 12x + c ____(1)

Given (2, -7) is a point on the curve.

(1) ⇒ -7 = (2)^{3} – 12(2) + c ⇒ -7 = 8 – 24 + c ⇒ c = 9

∴ Curve is y = x^{3} – 12x + 9.

Question 7.

Consider the curve x^{2/3} + y^{2/3} = 2

- Find the slope of the tangent to the curve at the point (1, 1). (2)
- Find the equation of the normal at the point (1, 1). (2)

Answer:

1. Given, x^{2/3} + y^{2/3} = 2,

Differentiating w.r.t. x,

⇒ slope of tangent = – 1.

2. Slope of normal = \(-\frac{1}{-1}\) = 1.

Equation of the normal is

y – 1 = 1(x – 1) ⇒ y – x = 0.

Question 8.

Find the intervals in which the function f given by f(x) = 2x^{3} – 3x^{2} – 36x + 7 is

- Strictly increasing. (2)
- Strictly decreasing. (2)

Answer:

Given; f(x) = 2x^{3} – 3x^{2} – 36x + 7

⇒ f'(x) = 6x^{2} – 6x – 36

f'(x) = 0 ⇒ 6x^{2} – 6x – 36 = 0

⇒ 6(x^{2} – x – 6) = 0

⇒ 6(x + 2)(x – 3) = 0 ⇒ x = -2, 3

The intervals are (-∞, -2),(- 2, 3), (3, ∞)

f'(-3) = 6(-3 + 2)(-3 – 3) > 0.

∴ Strictly increasing in (-∞, -2).

f'(0) = 6(2)(-3) < 0.

∴ Strictly decreasing in (- 2, 3).

f'(4) = 6(4 + 2)(4 – 3) > 0

∴ Strictly increasing in (3, ∞).

Question 9.

Use differentials to find the approximate value of \(\sqrt{0.6}\) up to 3 places of decimals.

Answer:

Take y = \(\sqrt{x}\), let x = 0.64 and ∆x = -0.04

Then; f(x) = y = \(\sqrt{x}\)

f(x + ∆x) = y + ∆y

(1) ⇒ \(\sqrt{0.6}\) = 0.8 – 0.025 = 0.775.

Question 10.

Use differentials to find the approximate value of (0.999)^{\(\frac{1}{10}\)} up to 3 places of decimals.

Answer:

Take = x^{\(\frac{1}{10}\)}, let x = 1 and ∆x = -0.001

Then; f(x) = y

f(x + ∆x) = y + ∆y

(0.999)^{\(\frac{1}{10}\)} = x^{\(\frac{1}{10}\)} + ∆y

(0.999)^{\(\frac{1}{10}\)} = (1)^{\(\frac{1}{10}\)} + ∆y

⇒ (0.999)^{\(\frac{1}{10}\)} = 1 + ∆y ______(1)

(1) ⇒ (0.999)^{\(\frac{1}{10}\)} = 1 – 0.0001 = 0.9999.

Question 11.

Use differentials to find the approximate value of (15)^{\(\frac{1}{4}\)} up to 3 places of decimals.

Answer:

Takey = x^{\(\frac{1}{4}\)}, let x = 16 and ∆x = -1

Then; f(x) = y

f(x + ∆x) = y + ∆y

(15)^{\(\frac{1}{4}\)} = x^{\(\frac{1}{4}\)} + ∆y

(15)^{\(\frac{1}{4}\)} = 16^{\(\frac{1}{4}\)} + ∆y

(15)^{\(\frac{1}{4}\)} = 2 + ∆y ____(1)

⇒ ∆y ≈ dy = \(\frac{d y}{d x}\) ∆x

Question 12.

Use differentials to find the approximate value of (26.57)^{\(\frac{1}{3}\)} up to 3 places of decimals.

Answer:

Take y = x^{\(\frac{1}{3}\)}, let x = 27 and ∆x = -0.43

Then; f(x) = y

f(x + ∆x) = y + ∆y

f(x + ∆x) = f(x) + ∆y

(26.57)^{\(\frac{1}{3}\)} = 27^{\(\frac{1}{3}\)} + ∆y .

⇒ (26.57)^{\(\frac{1}{3}\)} = 3 + ∆y ____(1)

(1) ⇒ (26.57)^{\(\frac{1}{3}\)} = 3 – 0.016 = 2.984.

Question 13.

Find the approximate value of f(5.001) where f(x) = x^{3} – 7x^{2} + 15

Answer:

Let x = 5 and ∆x = 0.001

Then; f(x) = y

f(x + ∆x) = y + ∆y

f (5.001) = f(x) + ∆y

f(5.001) = f(5) + ∆y

f(5.001) = 5^{3} – 7(5)^{2} + 15 + ∆y

⇒ f(5.001) = -35 + ∆y …

⇒ ∆y ≈ dy = \(\frac{d y}{d x}\) ∆x ⇒ dy = (3x^{2} – 14x) × 0.001

= (3(5)^{2} – 14(5)) × 0.001 = (75-70)0.001 = 0.005

(1) ⇒ f(5.001) = -35 + 0.005 = -34.995.

Question 14.

Find the approximate value of f(3.02) where f(x) = 3x^{2} + 5x + 3

Answer:

Let x = 3 and ∆x = 0.02

Then; f(x) = y

f(x + ∆x) = y + ∆y

f(3.02) = f(x) + ∆y

f(3.02) = f(3) + ∆y

f(3.02) = 3(3)^{2} + 5(2) + 3 + ∆y

⇒ f(3.02) = 45 + ∆y ____(1)

⇒ ∆y ≈ dy = \(\frac{d y}{d x}\) ∆x ⇒ dy = (6x + 5) × 0.02

= (6(3) + 5) × 0.02 = (18 + 5)0.02 = 0.46

(1) ⇒ f(3.02) = 45 + 0.46 = 45.46.

Question 15.

Consider the function y = f\(\sqrt{x}\)

- If x = 0.0036 and ∆x = 0.0001 find ∆y. (3)
- Hence approximate \(\sqrt{.0037}\) using differentials. (1)

Answer:

1. Let x = .0036, ∆x = 0.0001

2. (1) ⇒ \(\sqrt{.0037}\) = .000833 + .06 = .060833.

Question 16.

Find the approximate value of \(\sqrt[3]{124}\).

Answer:

f(x) = \(\sqrt[3]{x}\) = x^{1/3} ⇒ f^{1}(x) = 1/3x^{-2/3} = \(\frac{1}{3 x^{2 / 3}}\)

Let x = 125, ∆x = -1

Then; f(x) = y

Question 17.

Find two numbers x and y such that their sum is 35 and the product is x^{2} y^{5} a maximum.

Answer:

Given; x + y = 35 ⇒ y = 35 – x

P = x^{2} y^{5} ⇒ P = x^{2}(35 – x)^{5}

⇒ p’ = 2x(3 5 – x)^{5} + x^{2} 5(35 – x)^{4}(-1)

⇒ P’ = x(35 – x)^{4}[2(35- x) – 5x]

⇒ p’ = x(35 – x)^{4}[70 – 7x]

⇒ p’ = 7x(35 – x)^{4}[10 – x]

⇒ p” = 7[x(3 5 – x)^{4} [-1] + x(10 – x)4(35 – x)^{3} (-1) + (35 – x)^{4}(10 – x)]

For turning points P’ = 0

⇒ 7x(35 – x)^{4}[10 – x] = 0

⇒ x = 0, 35, 10

x = 0, 35 can be rejected since correspondingly y will be y = 35, 0

⇒ P” = 7[10(35 – 10)^{4}[-1]] < 0

Therefore maximum at x = 10

Thus the numbers are 10 and 35 – 10= 10.

Question 18.

Using differentials, find the approximate value of (63)^{1/3}.

Answer:

Take y = x^{\(\frac{1}{3}\)}, let x = 64 and ∆x = 1

Then; f(x) = y

f(x + ∆x) = y + ∆y

Question 19.

- Find the point on the curve y = x
^{3}– 10x + 8 at which the tangent is parallel to the line y = 2x + 1. (2) - Is the given line tangent to the curve? Why?

Answer:

1. \(\frac{d y}{d x}\) = 3x^{2} – 10

Slope of the line y = 2x +1 is 2

⇒ 3x^{2} – 10 = 2 ⇒ 3x^{2} = 12 ⇒ x = ±2

When x = 2

y = 2^{3} – 10 × 2 + 8 = 8 – 20 + 8 = -4

When x = – 2

y = (-2)^{3} -10 × (-2) + 8 = -8 + 20 + 8 = 20

Therefore the points are (2, -4); (-2, 20)

2. No. Since (2, -4); (-2, 20) does not satisfies the equation y = 2x +1.

Question 20.

Suppose that a spherical balloon is inflated and it has volume ‘v’ and radius ‘r’ at time ‘t’.

- If the balloon is inflated by pumping 900c.c. of gas per second. Find the rate a which the radius of the balloon is increasing when the radius is 15 cm. (2)
- Find the rate of change of its surface at the instant when it radius is 15 cm. (2)

Answer:

1. Let V be the volume of the sphere of radius r.

V = \(\frac{4}{3}\) πr^{3}, given; \(\frac{d V}{d t}\) = 900, r = 15

Differentiating w.r.t t,

2. Let ‘s’ denote the surface area of the balloon, then

S = 4πr^{2}

Differentiating, \(\frac{d s}{d t}\) = 4π.2r.\(\frac{d r}{d t}\) =8.π r .\(\frac{d r}{d t}\)

= 8π × 15 × \(\frac{1}{\pi}\) = 120 cm^{2}/sec.

Question 21.

Use differentials to find the approximate value of (0.009)^{\(\frac{1}{3}\)} up to 3 places of decimals.

Answer:

Take y = x^{\(\frac{1}{3}\)}, let x = 0.008 and ∆x = 0.001

Then; F(x) = Y

f(x + ∆x) = y + ∆y

Question 22.

Find the approximate value of \(\sqrt{401}\).

Answer:

f(x) = \(\sqrt{x}\) = x^{1/2}

f'(x) = \(\frac{1}{2 \sqrt{x}}\)

Let x = 400 ∆x = 1

f(x) = y = \(\sqrt{x}\)

f(x + ∆x) = y + ∆y

Question 23.

Consider y = \(\frac{\log x}{x}\), in (0, ∞)

- Find the value of x at which \(\frac{d y}{d x}\) = 0 (2)
- Find the maximum value.

Answer:

1.

2.

∴ y is maximum when x = e.

The maximum value is \(\frac{1}{e}\).

Question 24.

Find the point on the curve y = x^{3} – 11x + 5 at which the tangent is y = x – 11.

Answer:

Slope of the line y = x – 11 is 1.

Given; y = x^{3} – 11x + 5 ⇒ \(\frac{d y}{d x}\) = 3x^{2} – 11 = 1

⇒ 3x^{2} = 12 ⇒ x = ±2

At x = 2, ⇒ y = x – 11 = 2- 11 = -9

⇒ (2, -9)

At x = -2, ⇒ y = x – 11 = -2 – 11 = -13

⇒ (-2, -13)

But the point (-2, -13) do not lie on the curve, hence the point is (2, -9).

Question 25.

Consider the curve y = x^{2} – 2x + 7

- Find the slope of the tangent of the curve at x = 2. (2)
- Write down the equation of the tangent at x =2. (2)

Answer:

1. Given, y = x^{2} – 2x + 7 ⇒ y’ = 2x – 2

(y’)_{x=2} = 2(2) – 2 = 2.

2. At x = 2 , y = 2^{2} – 2(2) + 7 = 7.

Equation of the tangent at (2, 7) is

y – 7 = 2(x – 2) ⇒ 2x – y + 3 = 0.

Question 26.

Find the absolute maximum value and minimum value of the following functions.

- f(x) = 2x
^{3}– 15x^{2}+ 36x + 1, x ∈ [1, 5] - f(x) = 12x
^{\(\frac{4}{3}\)}– 6x^{\(\frac{1}{3}\)}, x ∈ [-1, 1]

Answer:

1. Given; f(x) = 2x^{3} – 15x^{2} + 36x + 1, x ∈ [1, 5]

⇒ f'(x) = 6x^{2} – 30x + 36

For turning point f'(x) = 0 ⇒ 6x^{2} – 30x + 36 = 0

⇒ x^{2} – 5x + 6 = 0 ⇒ (x -3)(x – 2) = 0

⇒ x = 3, 2

f(1) = 2(1)^{3} – 15(1)^{2} + 36(1) + 1 = 24

f(2) = 2(2)^{3} – 15(2)^{2} + 36(2) + 1 = 29

f(3) = 2(3)^{3} – 15(3)^{2} + 36(3) + 1 = 28

f(5) = 2(5)^{3} – 15(5)^{2} + 36(5) + 1 = 56

Absolute maximum = max {24, 29, 28, 56} = 56

Absolute minimum = min {24, 29, 28, 56} = 24

2. Given;

f'(x) = 0 at x = \(\frac{1}{8}\) and f'(x) is not defined at x = 0. Therefore;

Absolute maximum = max {18, 0, 6, \(-\frac{9}{4}\)} = 18

Absolute minimum = min {18, 0, 6, \(-\frac{9}{4}\)} = \(-\frac{9}{4}\).

Question 27.

Consider the function y = x^{3} – 6x^{2} + 3x – 1

- Find the slope at x= -1. (1)
- Find the minimum gradient of the above curve. (3)

Answer:

1. Given,

y = x^{3} – 6x^{2} + 3x – 1 ⇒ y’ = 3x^{2} – 12x + 3

Gradient at (x = -1) = (y’)_{x=1} = 3(-1)^{2} – 12(-1) + 3 = 18.

2. Now for minimum gradient we have to apply maxima – minima condition to the function y’ .ie, y” = 6x – 12 , for turning points of y’ is given by y” = 0.

Therefore, 6x – 12 = 0 ⇒ x = 2

Now, y”’ = 6 > 0

∴ y’ is maximum at x = 2.

Minimum gradient at (x = 2) is

= 12 – 24 + 3 = – 9.

### Plus Two Maths Application of Derivatives Six Mark Questions and Answers

Question 1.

A curve passes through the origin, and its gradient function is 2x – \(\frac{x^{2}}{2}\)

- Find its y coordinate when x= 2. (4)
- Find the equation of the tangent at x= 2. (2)

Answer:

Given;

Integrating we have; ∫dy = ∫(2x – \(\frac{x^{2}}{2}\))dx

⇒ y = x^{2} – \(\frac{x^{3}}{6}\) + c ___(1)

Since the curve passes through (0, 0)

(1) ⇒ 0 = 0 + c ⇒ c = 0

∴ Equation of the curve is y = x^{2} – \(\frac{x^{3}}{6}\)

When x = 2 ⇒ y = 2^{2} – \(\frac{2^{3}}{6}\) = \(\frac{8}{3}\)

∴ coordinate is (2, \(\frac{8}{3}\)).

2. Slope at (2, \(\frac{8}{3}\)) = 2 × 2 – \(\frac{2^{2}}{2}\) = 2

∴ Equation of the tangent at (2, \(\frac{8}{3}\)) is given by

y – \(\frac{8}{3}\) = 2(x – 2) ⇒ 3y – 8 = 6x – 12 ⇒ 3y = 6x – 4.

Question 2.

(i) Choose the correct answer from the bracket. The slope of the tangent to the curve y = x^{3} – 2x + 3 at x = 1 is ____(1)

(a) 0

(b) 1

(c) 2

(d) 3

(ii) Find points on the curve \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1 at which the tangents are (2)

(a) Parallel to x-axis

(b) parallel to y – axis.

(iii) Use differential to approximate \(\sqrt{25.6}\). (3)

Answer:

(i) (b) 1, Since

(ii)

(a) \(\frac{d y}{d x}\) = 0, since tangents are parallel to x- axis.

\(\frac{-9x}{25}\) = 0, x = 0 ∴ y = ± 3;

The points are (0, 3) and (0, -3)

(b) \(\frac{-25 y}{9 x}\) = 0, since tangents are parallel to y-axis, slope of normal = 0; y = o

∴ x = ± 5

The points are (5, 0) and (-5, 0)

(iii) Take y = \(\sqrt{x}\) , let x = 25 and ∆x = 0.6

Then; f(x) = y = \(\sqrt{x}\)

f(x + ∆x) = y + ∆y

Question 3.

Let x and y be the length and breadth of the rectangle ABCD in a circle having radius r. Let ∠CAB = θ (Ref. figure). If ∆ represent area of the rectangle and r is a constant.

- Write ∆ in terms of r and θ. (2)
- Find \(\frac{d \Delta}{d \theta}\) and \(\frac{d^{2} \Delta}{d \theta^{2}}\). (1)
- Hence find the maximum value of ∆. (2)
- Show that the rectangle of maximum area that can be inscribed in a circle of radius r is a square of side \(\sqrt{2} r\). (1)

Answer:

1. Area of the rectangle is ∆ = xy

From the figure y = 2r sinθ, x = 2r cosθ

∆ = xy = 4r^{2}sinθcosθ = 2r^{2}sin2θ

2. \(\frac{d \Delta}{d \theta}\) = 4r^{2} cos2θ ⇒ \(\frac{d^{2} \Delta}{d \theta^{2}}\) = -8r^{2}sin2θ

3. For turning points

Therefore local maximum at θ = \(\frac{\pi}{4}\)

4. Then;

Hence the rectangle becomes a square.

Question 4.

The second derivative of the equation of a curve is given by the equation x \(\frac{d^{2} y}{d x^{2}}\) = 1, given y = 1, \(\frac{d y}{d x}\) = 0 when x= 1.

- Find the slope at x = e. (2)
- Find the equation of the curve. (2)
- Find the equation of the normal at x= e. (2)

Answer:

1. Given;

Integrating we get,

∴ Slope of the curve at x = e is given by

\(\left(\frac{d y}{d x}\right)_{x=e}\) = loge = 1.

2. We have,

\(\frac{d y}{d x}\) = logx, ⇒ dy = logxdx

Integrating we get,

∫dy = ∫logx dx ⇒ y = logx.x – ∫\(\frac{1}{x}\).x dx + c_{2}

⇒ y = xlogx – x + c_{2} ____(2)

Given; y = 1 when x = 1

(2) ⇒ 1 = 1log1 – 1 + c_{2} ⇒ 1 = 0 – 1 + c_{2} ⇒ c_{2} = 2

Therefore the equation of the curve is

y = xlogx – x + 2

3. We have, y = xlogx – x + 2

When x = e

⇒ y = e log e – e + 2 ⇒ y = e – e + 2 = 2

So we have to find the slope at (e, 2),

We know; \(\left(\frac{d y}{d x}\right)_{x=e}\) = log e = 1

∴ Slope of the normal at (e, 2)= -1

∴ Equation of the normal at (e, 2) is.

y – 2 = (-1) (x – e)

y – 2 = – x + e ⇒ y + x = e + 2.

Question 5.

The given figure represents a cylinder Inscribed in a sphere.

- Find an expression for the volume V of the cylinder. (2)
- Find the height of the cylinder when its volume V is maximum. (2)
- Find the volume and radius of the largest cylinder. (2)

Answer:

1. From the right triangle ∆OAB,

y^{2} = R^{2} – x^{2} ⇒ y = \(\sqrt{R^{2}-x^{2}}\)

Which is the radius of the cylinder

Also height = 2 x

∴ Volume = V= π y^{2} × 2x = 2π(R^{2} – x^{2})x = 2π(R^{2}x – x^{3}).

2. Now,

For maximum or minimum,

3. Base radius =

Question 6.

If f(x) = x^{3} + 3x^{2} – 9x + 4 is a real function

- Find the intervals in which the function is increasing or decreasing. (3)
- Find the points of local maxima or local minima of f(x) (2)
- Graph of a function is given in the following figure:

Which among the following represents the graph of its derivative? (1)

Answer:

1. f'(x) = 3x^{2} + 6x – 9

For turning points f'(x) = 3x^{2} + 6x – 9 = 0

⇒ x = 1, -3

These turning point divide the domain of f(x) in the following intervals. (-∞, -3), (-3, 1), (1, ∞) in (-∞, -3)

⇒ f'(-4) = 3(-4)^{2} + 6(-4) – 9 > 0

Hence increasing.

In (-3, 1) ⇒ f'(0) = 3(0)^{2} + 6(0) – 9 < 0

Hence decreasing.

In (1, ∞) ⇒ f'(2) = 3(2)^{2} + 6(2) – 9 > 0

Hence increasing.

2. x = -3 is a local maximum point and x = 1 is a local minimum point.

3. (a)

The function passes through origin and has a local maximum at x = 2.

Question 7.

Of all the Cylinders with given surface area, show that the volume is maximum when height is equal to the diameter of the base.

Answer:

Let r be the radius, h be the height, V be the volume and S be the surface area

S = 2πr^{2} + 2 πrh

S – 6πr^{2} = 0

2πr^{2} + 2πrh – 6πr^{2} = 0

So h = 2r

So volume is maximum when h = 2r.

Question 8.

Sand is pouring from a pipe. The falling sand forms a Cone on the ground in such a way that the height of the Cone is always one-sixth of the radius of the base.

- Establish a relation between the volume ‘v’ and height ‘h’ of the Cone using the given condition. (2)
- lf the sand is pouring at the rate of -12 cm/sec, Find the rate of change of height of the Cone. (2)
- Find \(\frac{\mathrm{dh}}{\mathrm{dt}}\) when h = 4cm. (2)

Answer:

1. Given that the height of the Cone is one-sixth of the radius of the base, then h = \(\frac{r}{6}\) ⇒ r = 6h

Then Volume V = \(\frac{1}{3}\) πr^{2}h = \(\frac{1}{3}\) π(6h)2.h

V = \(\frac{1}{3}\) π 36h^{2}.h = \(\frac{1}{3}\) π 36h^{3}

V =12 πh^{3} ____(1)

2. Differentiating (1) we get

3. When h = 4 cm

Question 9.

(i) Choose the correct answer from the bracket. The rate of change of the area of a circle with respect to its radius r at r = 10cm is.

(a) 10π

(b) 20π

(c) 30π

(d) 40π (1)

(ii) Find the intervals in which the function f given by f(x) = x^{2} – 6x + 5 is (2)

(a) Strictly increasing

(b) Strictly decreasing

(iii) Find the local minimum and local maximum value, if any, of the function f(x) = x^{3} – 6x^{2} + 9x + 8 (3)

Answer:

(ii) f'(x) = 2x — 6; 2x – 6 = 0; x = 3

(-∞, 3 ) is strictly decreasing

(3, ∞) is strictly increasing.

(iii) f'(x) = 3x^{2} – 12x + 9

f^{11} = 6x — 12

For maxima, minima

f^{1} = 0 → 3x^{2} – 12x + 9 = 0

3(x – 3)(x — 1) = 0; x = 3, x = 1

At x = 3 f^{11}(x) = 6 × 3 – 12 = 18 – 12 = 6 > 0

f is minimum, the local minimum value of f = 8

At x = 1 f^{11}(x) = 6 × 1 – 12 = -6 < 0,

f is maximum, the local maximum value of f = 12.

Question 10.

A wire of length 28m is cut into two pieces. One of the pieces is be made into a square and the other into a circle. What should be the length of the two pieces so that combined area of the square and the circle is minimum using differentiation?

Answer:

Let the length of one piece be ‘x’ and other piece be ‘28 – x’. Let from the first piece we will make a circle of radius Y and from the second piece we will make a square of side y. Then,

Let A be the combined area of the circle and square, then

Question 11.

An open box of maximum volume is to be made from a square piece of tin sheet 24cm on a side by cutting equal squares from the corners and turning of the sides.

(i) Complete the following table. (2)

(ii) Using the above table, express V as a function of x and determine its domain. (1)

(iii) Find height (x. cm) of the box when volume V is maximum by differentiation. (3)

Answer:

(i)

(ii) Generalise the above table as a function.

V = x(24-2x)^{2}, 0 < x < 12.

(iii) \(\frac{d V}{d x}\) = x.2(24 – 2x)(-2) + (24 – 2x)^{2}

= -4x(24 – 2x) + (24 – 2x)^{2}

= -96x + 8x^{2} + 576 + 4x^{2} – 96x

= 12x^{2} – 192x + 576

For maximum or minimum,

Therefore volume is maximum when x = 4 cm.

Question 12.

A square tank of capacity 250 m^{3} has to be dug out. The cost of land is Rs. 50 per m^{2}. The cost of digging increases with the depth and for the whole tank is Rs. 400 × (depth)^{2}.

- Find an expression for the cost of digging the tank. (3)
- Find the dimension of the tank when the total cost is least. (3)

Answer:

1. Let x, x and y be the length, breadth, and depth of the tank.

Then, V = x. x. y = 250 ⇒ y = \(\frac{250}{x^{2}}\).

Area of land = x^{2}

⇒ Cost of land = 50 x^{2}

(∵ cost of land is Rs.50/m^{2})

Cost of digging = 400 × (depth)^{2} = 400 × (y)^{2}

∴ Total cost = C = 50 x^{2} + 400 × (y)^{2}

2. We have, C = 50x^{2} + \(\frac{400 \times(250)^{2}}{x^{4}}\).

Differentiating w.r.t.x, we get,

∴ Maximum at x = 10

m ⇒ when x = 10m and

y = \(\frac{250}{10^{2}}\) = 2.5m the total cost is least.

Question 13.

Show that the right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt{2}\) times the radius of the base.

Answer:

Volume of the cone will be, V =\(\frac{1}{3}\)πr^{2}h

h = \(\frac{3 V}{\pi r^{2}}\) ____(1)

Curved surface area will be, S = πrl

⇒ S^{2} = π^{2}r^{2}l^{2} = P

⇒ P = π^{2}r^{2}(h^{2} + r^{2}) ⇒ P = π^{2}r^{2}h^{2} + π^{2}r^{4})

⇒ 2r^{2} = h^{2} ⇒ h = \(\sqrt{2} r\).

Question 14.

Let ABC be an isosceles triangle inscribed in a circle having radius r. Then by figure, area of the triangle ABC is ∆

- Find \(\frac{d \Delta}{d \theta}\) and \(\frac{d^{2} \Delta}{d \theta^{2}}\) (2)
- Find the maximum value of ∆. (3)
- Show that the isosceles triangle of maximum area that can be in scribed in a given circle is an equilateral triangle. (1)

Answer:

1. Area of the isosceles triangle is ∆ = \(\frac{1}{2}\)bh

From the figure b = r sin2θ, h = r + r cos2θ

2. For turning points

Therefore local maximum a θ = \(\frac{\pi}{6}\) which means the area of the isosceles triangle is maximum When θ = \(\frac{\pi}{6}\).

3. Then; ∠OCB = 30° ⇒ ∠ACB = 2∠OCB = 60°. Therefore the isosceles triangle is an equilateral triangle.

Question 15.

(i) Using the graph of the function f (x) in the interval [ a, h ] match the following.

A – Point | B – Nature |

x = a | Absolute maximum |

x = b | Absolute minimum |

x = e | Local maximum |

x = d | Local minimum |

Point of inflexion. |

(ii) Consider the function f(x) = 3x^{4} – 8x^{3} + 12x^{2} – 48x + 25

(a) Find the turning points of f(x). (1)

(b) Explain the nature of the turning points (1)

(c) Find the absolute extreme values of f(x). (2)

Answer:

(i)

A – Point | B – Nature |

x = a | Absolute minimum |

x = b | Local maximum |

x = e | Point of inflexion |

x = d | Absolute maximum |

(ii) (a) f(x) = 12x^{3} – 24x^{2} + 24x – 48

For turning points,

f'(x) = 0 ⇒ 12x^{3} – 24x^{2} + 24x – 48 = 0

⇒ x^{3} – 2x^{2} + 2x – 4 = 0

⇒ (x^{2} + 2)(x – 2) = 0 ⇒ x = ± (\(\sqrt{-2}\), 2)

We admit only x = 2 as x = \(\sqrt{-2}\) is not a real number.

Therefore at x = 2 f (x) has a turning point.

(b) f”(x) = 36x^{2} – 48x + 24

⇒ f”(2) = 36(2)^{2} – 48 × 2 + 24 > 0

Therefore at x = 2 f(x) has a local minimum.

(c) f(0) = 25,

f(2) = 3(2)^{4} – 8(2)^{3} + 12(2)^{2} – 48 × 2 + 25 = -39

f(3) = 3(3)^{4} – 8(3)^{3} + 12(3)^{2} – 48 × 3 + 25 = 16

Consider the set { f (0), f{2), f (3)}

⇒ {25, -39, 16}

The maximum value of the above set is the absolute maximum and it is 25 at x = 0. The minimum value of the above set is the absolute minimum and it is -39 at x = 2.

Question 16.

An open box with a square base is to be made out of a given quantity of sheet of area a^{2}.

- If the box has side x units, then show that volume V= \(\frac{a^{2} x-x^{3}}{4}\) (2)
- Show that the maximum volume is \(\frac{a^{3}}{6 \sqrt{3}}\) (4)

Answer:

1. Area = a^{2} = x^{2} + 4xh,

h = height of the box.

2. We have,

Question 17.

For the function f(x) = sin2x, 0 < x < π

(i) Find the point between 0 and π that satisfies f'(x) = 0. (2)

(ii) Find the point of local maxima and local minima. (2)

(iii) Find the local maximum and local minimum value. (2)

Answer:

Question 18.

A cylindrical can with a volume of 125m3 (about 2 litres) is to be made by cutting its top and bottom from metal squares and forming its curved side by bending a rectangular sheet of metal to match its ends. What radius ‘r’ and height ‘h’ of the can will minimize the amount of material required.

Answer:

The circular top and bottom should be cut out from a square metal sheet of side 2r. Therefore the area of squares is 8r^{2}.

Area A = 8r^{2} + 2πrh

∴ To minimize the amount of material, r = 2.5

\(h=\frac{125}{\pi(2.5)^{2}}=6.3\).

Question 19.

A rectangle sheet of tin with adjascent sides 45cm and 24cm is to be made into a box • without top, by cutting off equal squares from the comers and folding up the flaps

- Taking the side of the square cut off as x, express the volume of the box as the function of x. (2)
- For what value of x, the volume of the box will be maximum. (4)

Answer:

1. Length of the box = 45 – 2x

Breadth of the box = 24 – 2x

Height of the box = x

Volume; V = (45 – 2x)(24 – 2x)x

= (1080 – 138x + 4x^{2})x

= 4x^{3} – 138x^{2} + 1080x.

2. \(\frac{d y}{d x}\) = 12x^{2} – 276x + 1080

12x^{2} – 276x + 1080 = 0

x^{2} – 23x + 90 = 0

x = 18, 5

x = 18 is impossible

∴ x = 5 when x = 5, \(\frac{d^{2} y}{d x^{2}}\) < 0

The volume of the box is maximum at x = 5.