CBSE Class 12 Mathematics Application of Derivatives Notes Set 03

These notes cover Chapter 6 Application of Derivatives of Class 12 Mathematics (CBSE): increasing and decreasing functions (monotonicity), local and absolute maxima and minima, and real-life optimisation problems. They include the important definitions and tests, solved examples with step marks, Do Yourself practice questions and their answers.

Monotonicity: Important Notes, Formulas and Solved Examples

1(A) Monotonicity at a Point

(I) A function f(x) is called an increasing function at point x = a, if in a sufficiently small neighbourhood of x = a; f(a − h) < f(a) < f(a + h)

(II) A function f(x) is called a decreasing function at point x = a, if in a sufficiently small neighbourhood of x = a; f(a − h) > f(a) > f(a + h)

Note: If x = a is a boundary point, then use the appropriate one sided inequality to test monotonicity of f(x).

(III) Testing of monotonicity of differentiable function at a point:
(i) If f′(a) > 0, then f(x) is increasing at x = a.
(ii) If f′(a) < 0, then f(x) is decreasing at x = a.
(iii) If f′(a) = 0, then examine the sign of \( f'(a^+) \) and \( f'(a^-) \).
(1) If \( f'(a^+) > 0 \) and \( f'(a^-) > 0 \), then increasing
(2) If \( f'(a^+) < 0 \) and \( f'(a^-) < 0 \), then decreasing
(3) otherwise neither increasing nor decreasing.

Illustration 1. Let f(x) = x³ − 3x + 2. Examine the nature of increasing and decreasing at points x = 0, 1 & 2. [4]
Answer:
f(x) = x³ − 3x + 2
f′(x) = 3(x² − 1) [1]

(i) f′(0) = −3 ⇒ decreasing at x = 0 [1]
(ii) f′(1) = 0
also, \( f'(1^+) \) = positive and \( f'(1^-) \) = negative
⇒ neither increasing nor decreasing at x = 1. [1]
(iii) f′(2) = 9 ⇒ increasing at x = 2 [1]

(B) Monotonicity over an Interval

(a) A function f(x) is said to be monotonically increasing (MI) in (a, b) if f′(x) ≥ 0 where equality holds only for discrete values of x i.e. f′(x) does not identically become zero for x ∈ (a, b) or any sub interval.
(b) f(x) is said to be monotonically decreasing (MD) in (a, b) if f′(x) ≤ 0 where equality holds only for discrete values of x i.e. f′(x) does not identically become zero for x ∈ (a, b) or any sub interval.
Note: A function is said to be monotonic if it is either increasing or decreasing.

Illustration 2. Determine for which values of x, the function \( y = x^4 - \frac{4x^3}{3} \) is increasing and for which values, it is decreasing. [Exemplar] [4]
Answer:
\( y = x^4 - \frac{4x^3}{3} \quad \Rightarrow \frac{dy}{dx} = 4x^3 - 4x^2 = 4x^2(x - 1) \)
Now, \( \frac{dy}{dx} = 0 \) ⇒ x = 0, x = 1. [2]
Since f′(x) < 0 ∀ x ∈ (−∞, 0) ∪ (0, 1) and f is continuous in (−∞, 0] and [0, 1].
Therefore f is decreasing in (−∞, 1] and f is increasing in [1, ∞).
Note: Here f is strictly decreasing in (−∞, 0) ∪ (0, 1) and is strictly increasing in (1, ∞) [2]

Illustration 3. Find the intervals in which \( f(x) = \frac{x}{2} + \frac{2}{x} \), x ≠ 0 is strictly increasing or decreasing. [4]
Answer:
We have,
\( f(x) = \frac{x}{2} + \frac{2}{x} \quad \Rightarrow f'(x) = \frac{1}{2} - \frac{2}{x^2} = \frac{x^2 - 4}{2x^2} \) [1]
For f(x) to be increasing, we must have
f′(x) > 0
\( \Rightarrow \frac{x^2 - 4}{2x^2} > 0 \Rightarrow x^2 - 4 > 0 \)
⇒ (x − 2)(x + 2) > 0 ⇒ x < −2 or x > 2 [1]
⇒ x ∈ (−∞, −2) ∪ (2, ∞) [½]

So, f(x) is increasing on (−∞, −2) ∪ (2, ∞).
For f(x) to be decreasing, we must have
f′(x) < 0 [½]
But, domain (f) = R − {0}.
So, f(x) is decreasing on (−2, 0) ∪ (0, 2). [1]

Illustration 4. Separate the interval [0, π/2] into sub-intervals in which f(x) = sin⁴x + cos⁴x is increasing or decreasing. [4]
Answer:
We have,
f(x) = sin⁴x + cos⁴x
⇒ f′(x) = 4 sin³x cos x − 4 cos³x sin x [½]
⇒ f′(x) = −4 sin x cos x (cos²x − sin²x) [½]
⇒ f′(x) = −2 (2 sin x cos x) (cos 2x)
⇒ f′(x) = −2 sin 2x cos 2x [½]
⇒ f′(x) = −sin 4x [½]
We have, 0 < x < π/2 ⇒ 0 < 4x < 2π.
Since sine function is positive in the first and second quadrants and negative in the third and fourth quadrants. So, we consider the following:
Case-I: When 0 < 4x < π
In this case, we have
sin 4x > 0
⇒ −sin 4x < 0 ⇒ f′(x) < 0
So, f(x) is decreasing on [0, π/4). [1]
Case-II: When π < 4x < 2π
In this case, we have
sin 4x < 0
⇒ −sin 4x > 0
⇒ f′(x) > 0
∴ f′(x) > 0 for π < 4x < 2π i.e. π/4 < x < π/2.
So, f(x) is increasing on (π/4, π/2] [1]

Do Yourself 1

(i) Find the values of k for which f(x) = kx³ − 9kx² + 9x + 3 is increasing on R.
(ii) Find the intervals in which function \( f(x) = \frac{x}{\log x} \) is increasing or decreasing.

Maxima and Minima: Important Notes, Formulas and Solved Examples

1. Introduction

Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. Here are examples of such problems that we will solve in this chapter
(i) What is the shape of a vessel that can withstand maximum pressure?
(ii) What is the maximum acceleration of a space shuttle? (This is an important question to the astronauts who have to withstand the effects of acceleration)
(iii) What is the radius of a contracted windpipe that expels air most rapidly during a cough?
These problems can be reduced to finding the maximum or minimum values of a function. Let's first explain exactly what we mean by maxima and minima.

(a) Maxima (Local/Relative maxima):
A function f(x) is said to have a maxima at x = a if there exist a neighbourhood (a − h, a + h) − {a} such that
f(a) > f(x) ∀ x ∈ (a − h, a + h) − {a}
(b) Minima (Local/Relative minima):
A function f(x) is said to have a minima at x = a if there exist a neighbourhood (a − h, a + h) − {a} such that f(a) < f(x) ∀ x ∈ (a − h, a + h) − {a}
(c) Absolute maxima (Global maxima):
A function f has an absolute maxima (or global maxima) at c if f(c) ≥ f(x) for all x in D, where D is the domain of f. The number f(c) is called the maximum value of f on D.
(d) Absolute minima (Global minima):
A function f has an absolute minima at c if f(c) ≤ f(x) for all x in D and the number f(c) is called the minimum value of f on D.

Note:
(i) the maximum & minimum values of a function are also known as local/relative maxima or local/relative minima as these are the greatest & least values of the function relative to some neighbourhood of the point in question.
(ii) the term 'extrema' is used both for maxima or minima.
(iii) a maximum (minimum) value of a function may not be the greatest (least) value in a finite interval.
(iv) a function can have several extreme values & a local minimum value may even be greater than a local maximum value.

2. Derivative Test for Ascertaining Maxima/Minima

(a) First derivative test:
If f′(x) = 0 at a point (say x = a) and
(i) If f′(x) changes sign from positive to negative in the neighbourhood of x = a then x = a is said to be a point local maxima.
(ii) If f′(x) changes sign from negative to positive in the neighbourhood of x = a then x = a is said to be a point local minima.

Note: If f′(x) does not change sign i.e. has the same sign in a certain complete neighbourhood of a, then f(x) is either increasing or decreasing throughout this neighbourhood implying that x = a is not a point of extremum of f. In this case x = a is a point of inflexion.

Illustration 5. Let \( f(x) = x + \frac{1}{x} \); x ≠ 0. Discuss the local maximum and local minimum values of f(x). Using the first derivative test. [2]
Answer:
Here, \( f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2} = \frac{(x - 1)(x + 1)}{x^2} \) [1]

Using number line rule, f(x) will have local maximum at x = −1 and local minimum at x = 1
∴ local maximum value of f(x) = −2 at x = −1
and local minimum value of f(x) = 2 at x = 1 [1]

Illustration 6. Find the local maxima or local minima, if any, of the function \( f(x) = \sin x + \cos x, \ 0 < x < \frac{\pi}{2} \) using the first derivative test. [4]
Answer:
We have,
y = f(x) = sin x + cos x ⇒ \( \frac{dy}{dx} = \cos x - \sin x \) [½]
For a local maximum or a local minimum, we have
\( \frac{dy}{dx} = 0 \Rightarrow \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4} \quad \left[\because 0 < x < \frac{\pi}{2}\right] \) [½]
Now, we will see whether \( x = \frac{\pi}{4} \) is a point of local maximum or a point of local minimum or none of these.
In the left nbd of \( x = \frac{\pi}{4} \), we have
\( 0 < x < \frac{\pi}{4} \Rightarrow \cos x > \sin x \Rightarrow \cos x - \sin x > 0 \Rightarrow \frac{dy}{dx} > 0 \) [1]
In the right nbd of \( x = \frac{\pi}{4} \), we have
\( \frac{\pi}{4} < x < \frac{\pi}{2} \Rightarrow \cos x < \sin x \Rightarrow \cos x - \sin x < 0 \Rightarrow \frac{dy}{dx} < 0 \) [1]
Thus, \( \frac{dy}{dx} \) changes its sign from positive to negative as x increases through \( \frac{\pi}{4} \).
So, f(x) attains a local maximum at \( x = \frac{\pi}{4} \). [1]

Illustration 7. Show that the function f(x) = 4x³ − 18x² + 27x − 7 has neither maxima nor minima. [Exemplar] [4]
Answer:
f(x) = 4x³ − 18x² + 27x − 7
f′(x) = 12x² − 36x + 27 = 3(4x² − 12x + 9) = 3(2x − 3)² [1]
f′(x) = 0 ⇒ \( x = \frac{3}{2} \) (critical point) [1]
Since f′(x) > 0 for all \( x < \frac{3}{2} \) and for all \( x > \frac{3}{2} \). [1]
Hence \( x = \frac{3}{2} \) is a point of inflexion i.e., neither a point of maxima nor a point of minima
\( x = \frac{3}{2} \) is the only critical point, and f has neither maxima nor minima. [1]

Do Yourself 2

(i) Find local maxima and local minima for the function f(x) = x³ − 3x using the first derivative test.
(ii) Find the local maximum or local minimum, if any, of the function \( f(x) = \sin^4 x + \cos^4 x, \ 0 < x < \frac{\pi}{2} \) using the first derivative test.

(b) Second derivative test:
If f(x) is continuous and differentiable at x = a where f′(a) = 0 and f″(a) also exists then for ascertaining maxima/minima at x = a, 2nd derivative test can be used -
(i) If f″(a) > 0 ⇒ x = a is a point of local minima
(ii) If f″(a) < 0 ⇒ x = a is a point of local maxima
(iii) If f″(a) = 0 ⇒ second derivative test fails. To identify maxima/minima at this point either first derivative test or higher derivative test can be used.

Illustration 8. If f(x) = 2x³ − 3x² − 36x + 6 has local maximum and minimum at x = a and x = b respectively, then find ordered pair (a, b) [4]
Answer:
f(x) = 2x³ − 3x² − 36x + 6
f′(x) = 6x² − 6x − 36 & f″(x) = 12x − 6 [1]
Now f′(x) = 0 ⇒ 6(x² − x − 6) = 0 ⇒ (x − 3)(x + 2) = 0 ⇒ x = −2, 3 [1]
f″(−2) = −30 < 0
∴ x = −2 is a point of local maximum [1]
f″(3) = 30 > 0
∴ x = 3 is a point of local minimum [1]
Hence, (−2, 3) is the required ordered pair.

Illustration 9. Find the point of local maxima of f(x) = sin x (1 + cos x) in x ∈ (0, π/2). [4]
Answer:
Let f(x) = sin x (1 + cos x) = \( \sin x + \frac{1}{2}\sin 2x \) [½]
⇒ f′(x) = cos x + cos 2x [½]
f″(x) = −sin x − 2 sin 2x [1]
Now f′(x) = 0 ⇒ cos x + cos 2x = 0
⇒ cos 2x = cos (π − x) ⇒ x = π/3 [1]
Also \( f''\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} - \sqrt{3} < 0 \) ∴ f(x) has a maxima at x = π/3 [1]

Illustration 10. Find the points of local maxima and local minima, if any, of the function. Find also the local maximum and local minimum values, if exist: \( f(x) = \sin 2x - x, \text{ where } -\frac{\pi}{2} \le x \le \frac{\pi}{2} \) [4]
Answer:
We have,
f(x) = sin 2x − x ⇒ f′(x) = 2 cos 2x − 1
For local maximum or minimum, we must have
f′(x) = 0
\( \Rightarrow 2\cos 2x - 1 = 0 \Rightarrow \cos 2x = \frac{1}{2} \)
\( \Rightarrow 2x = -\frac{\pi}{3} \text{ or } 2x = \frac{\pi}{3} \quad \left[\because -\frac{\pi}{2} \le x \le \frac{\pi}{2} \ \therefore -\pi \le 2x \le \pi\right] \)
\( \Rightarrow x = -\frac{\pi}{6} \text{ or } x = \frac{\pi}{6} \) [1]
Thus, \( x = -\frac{\pi}{6} \) and \( x = +\frac{\pi}{6} \) are possible points of local maxima or minima.
Now, we test the function at each of these points.
We have, f″(x) = −4 sin 2x.
At x = −π/6: We have,
\( f''\left(-\frac{\pi}{6}\right) = -4\sin\left(-\frac{\pi}{3}\right) = -4 \times \frac{-\sqrt{3}}{2} = 2\sqrt{3} > 0 \)
So, \( x = -\frac{\pi}{6} \) is a point of local minimum. [1]
The local minimum value is \( f\left(-\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{3}\right) + \frac{\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{\pi}{6} \) [½]
At x = π/6: we have,
\( f''\left(\frac{\pi}{6}\right) = -4\sin\frac{\pi}{3} = -4\left(\frac{\sqrt{3}}{2}\right) = -2\sqrt{3} < 0 \)
So, \( x = \frac{\pi}{6} \) is a point of local maximum. [1]
The local maximum value is \( f\left(\frac{\pi}{6}\right) = \sin\frac{\pi}{3} - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \) [½]

Do Yourself 3

(i) Find local maximum value of function \( f(x) = \frac{\ln x}{x} \)
(ii) Show that the local maximum of \( \left(\frac{1}{x}\right)^x \) exist at \( x = \frac{1}{e} \)

(c) Absolute Maximum and Minimum Values

Absolute max (min) value of any continuous function on [a, b] exist at any one of local max [min] or at boundary points.

Illustration 11. Find the maximum and minimum values of f(x) = x + sin 2x in the interval [0, 2π]. [4]
Answer:
We have,
f(x) = x + sin 2x ⇒ f′(x) = 1 + 2 cos 2x
f′(x) = 0 ⇒ 1 + 2 cos 2x = 0 [1]
\( \cos 2x = -\frac{1}{2} = \cos\frac{2\pi}{3} \)
\( \Rightarrow 2x = 2n\pi \pm \frac{2\pi}{3} \)
\( \Rightarrow x = n\pi \pm \frac{\pi}{3} \)
\( \Rightarrow x = \frac{\pi}{3}, \ x = \frac{2\pi}{3}, \ x = \frac{4\pi}{3}, \ x = \frac{5\pi}{3} \) [1]
Now, f(0) = 0 + sin 0 = 0,
\( f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\frac{2\pi}{3} = \frac{\pi}{3} + \frac{\sqrt{3}}{2}; \quad f\left(\frac{4\pi}{3}\right) = \frac{4\pi}{3} + \sin\frac{8\pi}{3} = \frac{4\pi}{3} + \frac{\sqrt{3}}{2} \)
\( f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} + \sin\frac{4\pi}{3} = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}; \quad f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sin\frac{10\pi}{3} = \frac{5\pi}{3} - \frac{\sqrt{3}}{2} \) [1]
and, f(2π) = 2π + sin 4π = 2π + 0 = 2π.
Of these values, the maximum value is 2π and the minimum value is 0.
Thus, the maximum value of f(x) is 2π and the minimum value is 0. [1]

Illustration 12. Find both the maximum and the minimum value of 3x⁴ − 8x³ + 12x² − 48x + 1 on the interval [1, 4]. [4]
Answer:
Let f(x) = 3x⁴ − 8x³ + 12x² − 48x + 1. Then,
f′(x) = 12x³ − 24x² + 24x − 48 and f″(x) = 36x² − 48x + 24 [1]
Now, f′(x) = 0
⇒ 12x³ − 24x² + 24x − 48 = 0
⇒ x³ − 2x² + 2x − 4 = 0
⇒ x²(x − 2) + 2(x − 2) = 0
⇒ (x − 2)(x² + 2) = 0
⇒ x = 2 [∵ x² + 2 ≠ 0] [1]
At x = 2, we have
f″(2) = 36(2)² − 48(2) + 24 = 72 > 0.
So, x = 2 is a point of local minimum. [½]
Now, f(2) = −63, f(1) = −40 and f(4) = 257. [½]
So, the minimum and maximum values of f(x) on [1, 4] are −63 and 257 respectively. [1]

Do Yourself 4

(i) Show that f(x) = sin 2x is maximum at \( x = \frac{\pi}{4} \) and \( \frac{5\pi}{4} \) in the interval [0, 2π].
(ii) Find the maximum value of 2x³ − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1].

3. Useful Formulae of Mensuration to Remember

(a) Volume of a cuboid = lbh.
(b) Surface area of a cuboid = 2(lb + bh + hl).
(c) Curved surface area of a cylinder = 2πrh.
(d) Total surface area of a cylinder = 2πrh + 2πr².
(e) Volume of cylinder = πr²h
(f) Curved surface area of cone = πrl.
(g) Total surface area of cone = πrl + πr²
(h) Volume of a cone = \( \frac{1}{3}\pi r^2 h \).
(i) Surface area of a sphere = 4πr².
(j) Volume of a sphere = \( \frac{4}{3}\pi r^3 \).
(k) Area of a circular sector = \( \frac{1}{2}r^2\theta \), when θ is in radians.

4. Summary of Working Rules for Solving Real Life Optimization Problem

First: When possible, draw a figure to illustrate the problem & label those parts that are important in the problem. Constants & variables should be clearly distinguished.
Second: Write an equation for the quantity that is to be maximized or minimized. If this quantity is denoted by 'y', it must be expressed in terms of a single independent variable x. This may require some algebraic manipulations.
Third: If y = f(x) is a quantity to maximize or minimize then find those values of x for which dy/dx = f′(x) = 0.
Fourth: Using derivative test, test each value of x for which f′(x) = 0 to determine whether it provides a maximum or minimum or neither.
Fifth: If the derivative fails to exist at some point, examine this point as possible maximum or minimum point.
Sixth: If the function y = f(x) is defined only for x ∈ [a, b] then examine x = a & x = b also for possible extreme values.

Illustration 13. Amongst all pairs of positive numbers with product 256, find those whose sum is the least. [6]
Answer:
Let the required numbers be x and y. Then,
xy = 256 (given) ..............(i) [½]
Let S = x + y. Then,
\( S = x + \frac{256}{x} \) [Using (i)]
\( \Rightarrow \frac{dS}{dx} = 1 - \frac{256}{x^2} \text{ and } \frac{d^2S}{dx^2} = \frac{512}{x^3} \) [1]
For maximum or minimum values of S, we must have
\( \frac{dS}{dx} = 0 \) [½]
\( \Rightarrow 1 - \frac{256}{x^2} = 0 \Rightarrow x^2 = 256 \Rightarrow x = 16 \) [1]
Now, \( \left(\frac{d^2S}{dx^2}\right)_{x=16} = \frac{512}{(16)^3} = \frac{1}{8} > 0 \) [1]
Thus, S is minimum when x = 16. [1]
Putting x = 16 in (i) we get y = 16.
Hence, the required numbers are both equal to 16. [1]

Illustration 14. A box of maximum volume with top open is to be made by cutting out four equal squares from four corners of a square tin sheet of side length a feet and then folding up the flaps. Find the side of the square base cut off. [6]
Answer:
Volume of the box is, V = x(a − 2x)² i.e., squares of side x are cut out then we will get a box with a square base of side (a − 2x) and height x. [2]

\( \therefore \frac{dV}{dx} = (a - 2x)^2 + x \cdot 2(a - 2x)(-2) \) [½]
\( \frac{dV}{dx} = (a - 2x)(a - 6x) \) [1]
For V to be extremum \( \frac{dV}{dx} = 0 \Rightarrow x = \frac{a}{2}, \frac{a}{6} \) [1]
But when x = a/2; V = 0 (minimum) and we know minimum and maximum occurs alternately in a continuous function. [1]
Hence, V is maximum when x = a/6. [½]

Illustration 15. Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius a is a square of side \( \sqrt{2}a \). [6]
Answer:
Let ABCD be a rectangle in a given circle of radius a with centre at O.
Let AB = 2x and AD = 2y be the sides of the rectangle. Then

AM² + OM² = OA² [∵ ΔOAM is a right angled triangle]
\( \Rightarrow x^2 + y^2 = a^2 \Rightarrow y = \sqrt{a^2 - x^2} \) ..........(i) [1]
Let P be the perimeter of the rectangle ABCD. Then
P = 4x + 4y
\( \Rightarrow P = 4x + 4\sqrt{a^2 - x^2} \) [Using (i)]
\( \Rightarrow \frac{dP}{dx} = 4 - \frac{4x}{\sqrt{a^2 - x^2}} \) [1]
For maximum or minimum values of P, we have
\( \frac{dP}{dx} = 0 \)
\( \Rightarrow 4 - \frac{4x}{\sqrt{a^2 - x^2}} = 0 \Rightarrow 4 = \frac{4x}{\sqrt{a^2 - x^2}} \)
\( \Rightarrow \sqrt{a^2 - x^2} = x \Rightarrow a^2 - x^2 = x^2 \Rightarrow 2x^2 = a^2 \Rightarrow x = \frac{a}{\sqrt{2}} \) [1]
Now, \( \frac{d^2P}{dx^2} = \frac{-4\left\{\sqrt{a^2 - x^2} \cdot 1 - \frac{x(-x)}{\sqrt{a^2 - x^2}}\right\}}{\left(\sqrt{a^2 - x^2}\right)^2} = \frac{-4a^2}{(a^2 - x^2)^{3/2}} \) [1]
\( \therefore \left(\frac{d^2P}{dx^2}\right)_{x = a/\sqrt{2}} = \frac{-4a^2}{\left(a^2 - \frac{a^2}{2}\right)^{3/2}} = \frac{-8\sqrt{2}}{a} < 0 \) [½]
Thus, P is maximum when \( x = \frac{a}{\sqrt{2}} \) [½]
Putting \( x = \frac{a}{\sqrt{2}} \) in (i), we obtain \( y = \frac{a}{\sqrt{2}} \). Therefore, \( x = y = \frac{a}{\sqrt{2}} \Rightarrow 2x = 2y \).
Hence, P is maximum when the rectangle is square of side \( 2x = \frac{2a}{\sqrt{2}} = \sqrt{2}a \). [1]

Illustration 16. If a right circular cylinder is inscribed in a given cone. Find the dimension of the cylinder such that its volume is maximum. [6]
Answer:
Let x be the radius of cylinder and y be its height
V = πx²y [½]

x, y can be related by using similar triangles ΔDEC and ΔABC
\( \frac{y}{r - x} = \frac{h}{r} \Rightarrow y = \frac{h}{r}(r - x) \) [1]
\( \Rightarrow V(x) = \pi x^2 \frac{h}{r}(r - x) \Rightarrow V(x) = \frac{\pi h}{r}(rx^2 - x^3) \)
\( V'(x) = \frac{\pi h}{r}(2rx - 3x^2) \) [1]
\( V'(x) = 0 \Rightarrow x = 0, \frac{2r}{3} \) [1]
\( V''(x) = \frac{\pi h}{r}(2r - 6x) \) [1]
V″(0) = 2πh > 0 ⇒ x = 0 is point of minima [½]
\( V''\left(\frac{2r}{3}\right) = -2\pi h < 0 \Rightarrow x = \frac{2r}{3} \) is point of maxima
Thus volume is maximum at \( x = \frac{2r}{3} \) and \( y = \frac{h}{3} \). [1]

Illustration 17. A figure consists of a semi-circle with a rectangle on its diameter. Given the perimeter of the figure, find its dimensions in order that the area may be maximum. [6]
Answer:
Let ABCD be a rectangle and let the semi-circle be described on side AB as diameter. Let AB = 2x and AD = 2y. Let P be the perimeter and A be the area of the figure. Then,

P = 2x + 4y + πx or 4y = P − 2x − πx (Given) ....(i) [1]
and, \( A = (2x)(2y) + \frac{\pi x^2}{2} \) ....(ii) [1]
\( \Rightarrow A = 4xy + \frac{\pi x^2}{2} \)
\( \Rightarrow A = x(P - 2x - \pi x) + \frac{\pi x^2}{2} \) [Using (i)]
\( \Rightarrow A = Px - 2x^2 - \pi x^2 + \frac{\pi x^2}{2} \) [½]
\( \Rightarrow A = Px - 2x^2 - \frac{\pi x^2}{2} \) [½]
Eqⁿ diffⁿ w.r.t. x
\( \Rightarrow \frac{dA}{dx} = P - 4x - \pi x \text{ and } \frac{d^2A}{dx^2} = -4 - \pi \)
For maximum or minimum A, we must have
\( \frac{dA}{dx} = 0 \Rightarrow P - 4x - \pi x = 0 \Rightarrow x = \frac{P}{\pi + 4} \) [1]
Clearly, \( \frac{d^2A}{dx^2} = -4 - \pi < 0 \) for all values of x. [1]
Thus, A is maximum when \( x = \frac{P}{\pi + 4} \).
Putting \( x = \frac{P}{\pi + 4} \) in (i) we get \( y = \frac{P}{2(\pi + 4)} \).
So, dimensions of the figure are \( 2x = \frac{2P}{\pi + 4} \) and \( 2y = \frac{P}{\pi + 4} \). [1]

Illustration 18. The three sides of a trapezium are equal each being 6 cm long; find the area of trapezium when it is maximum. [6]
Answer:
Let ABCD be the given trapezium.

Let AM = BN = x cm [½]
then \( DM = CN = \sqrt{36 - x^2} \) [½]
∴ Area of trapezium ABCD is
\( S = \frac{1}{2}(6 + x + 6 + x) \times \sqrt{36 - x^2} \) [1]
\( = (6 + x)\sqrt{36 - x^2} \)
or S² = (6 + x)²(36 − x²)
Let y = (6 + x)²(36 − x²) [½]
\( \therefore \frac{dy}{dx} = (6 + x)^2(-2x) + (36 - x^2) \cdot 2(6 + x) \)
\( = 2(6 + x)^2(6 - 2x) = 4(3 - x)(6 + x)^2 \) [1]
and \( \frac{d^2y}{dx^2} = -12x(6 + x) \)
For max. or min. of y, \( \frac{dy}{dx} = 0 \) then x = 3 (x ≠ −6) [1]
\( \therefore \left.\frac{d^2y}{dx^2}\right|_{x=3} = -324 < 0 \) [½]
∴ y is maximum at x = 3 then S is also maximum at x = 3
\( \therefore S = (6 + 3)\sqrt{36 - 9} = 27\sqrt{3}\ \text{cm}^2 \) [1]

Illustration 19. An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. Show that the area of triangle is maximum when \( \theta = \frac{\pi}{6} \). [Exemplar] [6]
Answer:
Let ABC be an isosceles triangle inscribed in the circle with radius a such that AB = AC.
AD = AO + OD = a + a cos 2θ and BC = 2BD = 2a sin 2θ (see figure)

Therefore, area of the triangle ABC i.e. \( \Delta = \frac{1}{2}BC \cdot AD \)
\( = \frac{1}{2} \cdot 2a\sin 2\theta \cdot (a + a\cos 2\theta) \)
\( = a^2 \sin 2\theta(1 + \cos 2\theta) \)
\( \Rightarrow \Delta = a^2 \sin 2\theta + \frac{1}{2}a^2 \sin 4\theta \) [2]
Therefore, \( \frac{d\Delta}{d\theta} = 2a^2 \cos 2\theta + 2a^2 \cos 4\theta = 2a^2(\cos 2\theta + \cos 4\theta) \)
\( \frac{d\Delta}{d\theta} = 0 \Rightarrow \cos 2\theta = -\cos 4\theta = \cos(\pi - 4\theta) \) [2]
Therefore, \( 2\theta = \pi - 4\theta \Rightarrow \theta = \frac{\pi}{6} \)
\( \frac{d^2\Delta}{d\theta^2} = 2a^2(-2\sin 2\theta - 4\sin 4\theta) < 0 \ \left(\text{at } \theta = \frac{\pi}{6}\right) \)
Therefore, Area of triangle is maximum when \( \theta = \frac{\pi}{6} \). [2]

Do Yourself 5

(i) A jet of an enemy is flying along the curve y = x² + 2. A soldier is placed at the point (3, 2). What is the nearest distance between the soldier and the jet.
(ii) Of all closed right circular cylinders of a given volume of 100 cubic centimetres, find the dimensions of cylinder which has minimum surface area.

Answers for Do Yourself

Do Yourself 1: (i) 0 ≤ k ≤ 1/3, i.e. k ∈ [0, 1/3] (ii) Increasing (e, ∞), Decreasing (0, e) − {1}
Do Yourself 2: (i) local max. at x = −1, local min. at x = 1 (ii) local minimum at \( x = \frac{\pi}{4} \), minimum value = \( \frac{1}{2} \)
Do Yourself 3: (i) \( \frac{1}{e} \)
Do Yourself 4: (ii) maximum value = 89 at x = 3 and maximum value = 139 at x = −2
Do Yourself 5: (i) \( \sqrt{5} \) (ii) \( r = \left(\frac{50}{\pi}\right)^{1/3} \) cm & \( h = 2\left(\frac{50}{\pi}\right)^{1/3} \) cm.