Practicing the AP Inter 2nd Year Maths Study Material and AP Inter 2nd Year Maths Chapter 6 Application of Derivatives Solutions Exercise 6c will help students to clear their doubts quickly.
Inter 2nd Year Maths Application of Derivatives Solutions Exercise 6c
Application of Derivatives Exercise 6c Solutions
I. Find the maximum and minimum values, if any, of the following functions is given by
Question 1.
f(x) = | x + 21 – 1
Solution:
Given that f(x) = |x + 2| – 1
We know that |x + 2| > 0 for every x ∈ R.
∴ f(x) = |x + 2| – 1 ≥ -1 for every x ∈ R.
The minimum value of f is attained when |x + 2| = 0
|x + 2| = 0 ⇒ x = -2
Minimum value of f = f(-2) = |-2 + 2| – 1 = -1
Hence, function f does not have a maximum value.
Question 2.
g(x) = – |x + 1| + 3
Solution:
Given that g(x) = – |x + 1| + 3
We know that |x + 1| > 0 for every x ∈ R.
g(x) = – |x + 1| + 3 < 3 for every x ∈ R
The maximum value of g is attained when |x + 1| = 0 ⇒ x = -1
Maximum value of g = g (-1) = -|-1 + 1| + 3 = 3
Hence function ‘g’ does not have a minimum value.
Question 3.
h(x) = sin (2x) + 5
Solution:
Given that h(x) = sin (2x) + 5
We know that – 1 ≤ sin 2x ≤ 1
-1 + 5 ≤ sin 2x + 5 ≤ 1 + 5 ⇒ 4 ≤ sin 2x + 5 ≤ 6
Hence, the maximum and minimum values of h are 6 and 4 respectively.
Question 4.
f(x) = |sin 4x + 3|
Solution:
Given that f(x) = |sin 4x + 3|
We know that -1 ≤ sin4x ≤ 1
⇒ -1 + 3 ≤ sin4x + 3 ≤ 1 + 3 ⇒ 2 ≤ |sin4x + 3| ≤ 4
Hence, the maximum and minimum values of f are 4 and 2 respectively.
Application of Derivatives Class 12 Solutions Exercise 6c
Question 5.
h(x) = x + 1, x ∈ (-1, 1)
Solution:
Given that h(x) = x + 1, x ∈ (-1, 1)
Here, if a point x0 is closest to -1 then we find
\(\frac{\mathrm{x}_0}{2}\) + 1 < x0 + 1 for all x0 ∈ (-1, 1)
Also, if x1 is closest to 1, then x1 + 1 < \(\frac{x_1+1}{2}\) + 1
for all x ∈ (-1, 1)
Hence, the function h(x) has neither maximum nor minimum values in (-1, 1).
Question 6.
Prove that the following functions do not have maxima or minima:
f(x) = ex
Solution:
Given that f (x) = ex ⇒ f ‘(x) = ex
Now, if f (x) = 0, then e = 0
But, the exponential function can never assume 0 for any value of x.
∴ There does not exist c ∈ R such that f (c) = 0.
Hence, the function f does not have maxima or minima.
Question 7.
g(x) = log x
Solution:
Given that, g(x) = log x
⇒ g’(x) = \(\frac{1}{x}\), since x is defined for a positive number x, g’(x) > 0 for any x.
∴ There does not exist c ∈ R such that g’(c) = O
Hence, the function g does not have maxíma or minima.
Question 8.
h(x) = x3 + x2 + x + 1
Solution:
Given that h(x) = x3 + x2 + x + 1
= h(x) = 3x2 + 2x + 1
Now, h'(x) = 0 ⇒ 3x2 + 2x + 1 = 0
x = \(\frac{-2 \pm 2 \sqrt{2} \mathrm{i}}{6}=\frac{-1 \pm \sqrt{2} \mathrm{i}}{3}\) ∉ R
∴ There does not exist c E R such that h’ (c) = O
Hence, the function f does not have maxima or minima.
Question 9.
It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.
Solution:
Let f(x) = x4 – 62x2 + ax + 9 ⇒ f ‘(x) = 4x3 – 124x + a
It is given that the function f attains its maximum value on the interval [0, 2] at x = 1.
f’(1) = 0 ⇒ 4 (1)3 – 124 (1) + a = 0
⇒ 4- 124 + a = 0 ⇒ a = 120
Hence, the value of a is 120.
II. Find the maximum and minimum values, if any, of the following functions given by
Question 1.
f(x) = (2x – 1)2 + 3
Solution:
The given function is f(x) = (2x – 1)2 + 3
It can be observed that (2x – 1)2 > 0 for every x ∈ R
∴ f(x) = (2x – 1)2 + 3 > 3 for every x ∈ R.
The minimum value of f is attained when
2x – 1= 0 ⇒ x = \(\frac{1}{2}\)
∴ The minimum value of f = f(\(\frac{1}{2}\)) = [2.(\(\frac{1}{2}\)) – 1]2 + 3 = 3
Hence, the function f does not have a maximum value.
Question 2.
f(x) = 9x2 + 12 + 2
Solution:
The given function is f(x) = 9x2 + 12x + 2
= (3x + 2)2 – 2
¡(can be observed that (3x + 2)2 ≥ 0 for even x ∈ R.
The minimum value of f is attained when 3x + 2 = 0
⇒ x = –\(\frac{2}{3}\)
∴ The minimum value of f is
f(-\(\frac{2}{3}\)) = (3.(\(\frac{-2}{2}\))+ 2)2 – 2 = -2
Hence, the function f does not have a maximum value,
Question 3.
f(x) = – (x – 1)2 + 10
Solution:
The given function is f(x) = – (x – 1)2 + 10
It can be observed that (x – 1)2 ≥ 0 for every x ∈ R.
∴ f(x) = -(x – 1)2 + 10 ≤ 10 for every x ∈ R.
∴ Maximum value of f is attained when
(x – 1) = 0 ⇒ x = 1
Hence, the function f does not have a minimum value.
Question 4.
g(x) = x3 + 1
Solution:
The given function is g(x) = x3 + 1
Here, x3 > 0 if x > 0 and x3 < 0 if x < 0
Hence, the function ‘g’ has neither a maximum value nor a minimum value.
Question 5.
Find the local maxima and local minima, if any, ! of the following functions. Find also the local maximum and the local minimum values, as the case may be:
f(x) = x2
Solution:
Given that f(x) = x2 ⇒ f ‘(x) = 2x
Now, f ‘(x) = 0 ⇒ x = 0
Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.
We have f ‘(0) = 0, which is positive.
By second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) = 0.
Question 6.
g(x) = x3 – 3x
Solution:
Given that g(x) = x3 – 3x => g'(x) = 3x2 – 3
Now, g'(x) = 0 ⇒ x = ± 1 and g”(x) = 6x
g”(1) = 6 > 0 and g”(-1) = – 6 > 0
By second derivative test, x = 1 is a point of local minima and local minimum value of g at x = 1 is
g(1) = 13 – 3 = 1 – 3 = -2
However, x = – 1 is a point of local maxima and local maximum value of g at x = – 1 is
g (-1) = (-1)3 – 3 (-1) = -1 + 3 = 2
Question 7.
f(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\)
Solution:
Given that f (x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\)
⇒ f ‘(x) = cos x + sin x
f ‘(0) = 0 ⇒ cos x = – sin x ⇒ tan x = – 1
⇒ x = \(\frac{\pi}{4}\) ∈ (0, \(\frac{\pi}{2}\))
h”(x) = -sin x – cos x = – (sin x + cos x)
h”(\(\frac{\pi}{4}\)) = \(-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=\frac{-2}{\sqrt{2}}\) = -√2 < 0
∴ By second derivative test, x = \(\frac{\pi}{4}\) is a point of local maxima and the local maximum value of h
at x = \(\frac{\pi}{4}\) is
h(\(\frac{\pi}{4}\)) = sin\(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\)
= √2
Question 8.
f(x) = sin x – cos x, 0 < x < 2π
Solution:
Given that f(x) = sin x – cos x, 0 < x < 2π
f’(x) = cos x + sin x
f ‘(0) = 0 ⇒ cos x = – sin x
⇒ tan x = – 1
⇒ x = \(\frac{1}{2}\) ∈ (0, 2π)
f’'(x) = -sin x + cos x
f”(\(\frac{3 \pi}{4}\)) = -sin\(\frac{3 \pi}{4}\) + cos\(\frac{3 \pi}{4}\)
= \(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\) = -√2 < 0
f”(\(\frac{7 \pi}{4}\)) = -sin\(\frac{7 \pi}{4}\) + cos \(\frac{7 \pi}{4}\)
= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2 < 0 ∴ By second derivative test x = \(\frac{3 \pi}{4}\) is a point of local maxima and local maximum value of f at x = \(\frac{3 \pi}{4}\) is f(\(\frac{3 \pi}{4}\) ) = sin\(\frac{3 \pi}{4}\) – cos \(\frac{3 \pi}{4}\) = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2 > 0
However, x = \(\frac{7 \pi}{4}\) is a point of local minima and local minimum value of f at x = \(\frac{7 \pi}{4}\) is
f(\(\frac{7 \pi}{4}\)) = sin\(\frac{7 \pi}{4}\) – cos\(\frac{7 \pi}{4}\)
= \(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\)
= -√2
Question 9.
f(x) = x3 – 6x2 + 9x+ 15.
Solution:
Given that f(x) = x3 – 6x2 + 9x + 15
f'(x) = 3x2 – 12x + 9
f ‘(x) = 0 ⇒ 3 (x2 – 4x + 3) = 0
⇒ 3(x – 1)(x – 3) = 0 ⇒ x = 1, 3
Now, f “(x) = 6x – 12 = 6 (x – 2)
f “(1) = 6 (1 – 2) = – 6 < 0 and f “(3) = 6 (3 – 2) = 6 > 0
By second derivative test, x = 1 is a point of local maxima and local maximum value of f at x = 1 is
f(1) = 1 – 6 + 9 + 15 = 19
However, x = 3 is a point of local minima and local minimum value of f at x = 3 is
f(3) = 27 – 54 + 27 + 15 = 15
Question 10.
g(x) = \(\frac{x}{2}+\frac{2}{x}\), x > 0
Solution:
Given that g(x) = \(\frac{x}{2}+\frac{2}{x}\), x > 0
⇒ g'(x) = \(\frac{1}{2}-\frac{2}{x^2}\)
Now, g'(x) = 0 ⇒ \(\frac{2}{x^2}=\frac{1}{2}\) ⇒ x2 = 4 ⇒ x = ± 2
∴ x > 0, we take x = 2
Now, g”(x) = \(\frac{4}{x^3}\) ⇒ g”(2) = \(\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}\) > 0
x > 0, we take x = 2
Now, g”(x) = \(\frac{1}{2}\) ⇒ g”(2) = \(\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}\) > 0
By second derivative test, x 2 is a point of minima and local minimum value of g at x = 2 is
g(2) = \(\frac{2}{2}+\frac{2}{2}\) = 1 + 1 = 2
Question 11.
g(x) = \(\frac{1}{x^2+2}\)
Solution:
Given that g(x) = \(\frac{1}{x^2+2}\) ⇒ g'(x) = \(\frac{(-2 x)}{\left(x^2+2\right)^2}\)
g'(x) = 0 ⇒ \(\frac{(-2 x)}{\left(x^2+2\right)^2}\) = 0 ⇒ x = 0
Now, for values close to x = 0 and to the left of 0, g'(x) > 0. Also, for values close to x = 0 and to the right of 0, g'(x) < 0.
∴ By second derivative test, x = 0 is a point of local maxima and local maximum value of g(0) is
\(\frac{1}{0+2}\) = \(\frac{1}{2}\)
Question 12.
f(x) = \(x \sqrt{1-x}\), 0 < x < 1
Solution:
Given that f(x) = \(x \sqrt{1-x}\)
∴ By second derivative test, x = \(\frac{2}{3}\) is a point of local maxima and local maximum value of f at x = \(\frac{2}{3}\) is
\(f\left(\frac{2}{3}\right)=\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}=\frac{2}{3 \sqrt{3}}=\frac{2 \sqrt{3}}{9}\)
Find the absolute maximum value and the absolute minimum value of the following function in the given intervals. (13, -16)
Question 13.
f (x) = x3 x ∈ [- 2, 2]
Solution:
Given f(x) = x3 ⇒ f'(x) = 3x2
Now f'(x) = 0 ⇒ 3x2 = 0
Then we evaluate the value of f at critical point x = 0 and at end points of the interval [-2, 2]
(i) f(0) = 0 ……………… (1)
(ii) f(-2) =(-2)3 = -8 ………… (2)
(iii) f(2) = (2)3 =8 ………….. (3)
So, we can conclude that the absolute maximum value of f on [-2, 2] is 8 at x = 2.
Also the absolute minimum value of f on [-2, 2] is -8 occurring at x = -2.
Question 14.
f(x) = sin x + cos x in the given interval x ∈ [0, π]
Solution:
Given f(x) = sin x + cos x
⇒ f'(x) = cos x – sin x
Now f'(x) = 0 ⇒ cos x – sin x = 0 ⇒ sin x = cos x – 1
⇒ x = \(\frac{\pi}{4}\) ∈ [0, π]
Then we evaluate the value of f at critical point x = \(\frac{\pi}{4}\) and at the end points of the interval [0, π].
(i) \(f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4} \Rightarrow=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\) ………….. (1)
(ii) f(0) = sin 0 + cos 0 = 0 + 1 = 1 ……. (2)
(iii) f(π) = sin π + cos π – 0 – 1 = -1 ………….. (3)
Hence, we can conclude that the absolute maximum value of f on [0, π] is \(\sqrt{2}\) occurring at x = \(\frac{\pi}{4}\) and the absolute minimum value of f on [0, π] is -1 occurring at x = π.
Question 15.
f(x) = 4x – \(\frac{1}{2}\)x2 in the given interval x ∈ [-2, \(\frac{9}{2}\)]
Solution:
Given f(x) = 4x – \(\frac{1}{2}\)x2 and f'(x) = 4 – \(\frac{1}{2}\)(2x) = 4 – x
Now f'(x) = 0 ⇒ 4 – x = 4
Then, we evaluate the value of f at critical point x = 4 and at the end points of the interval [2, \(\frac{9}{2}\)]
(i) f(4) = 16 – \(\frac{1}{2}\)(16) = 16 – 8 = 8 …………… (1)
(ii) f(-2) = -8 – \(\frac{1}{2}\) (4) = -8 – 2 = -10 ……………. (2)
(iii) f\(\left(\frac{9}{2}\right)=\) = 18 – \(\frac{1}{2}\left(\frac{9}{2}\right)^2\) = 18 – \(\frac{81}{8}\) = 18 – 10.125 = 7.875 …………… (3)
Hence, we can conclude that the absolute maximum value of f on [-2, \(\frac{9}{2}\)] is 8 occurring at x = 4 and the absolute minimum value of f on [-2, \(\frac{9}{2}\)] is -10 occurring at x = -2.
Question 16.
f (x) = (x – 1)2 + 3, x ∈ [-3, 1]
Solution:
Given f(x) = (x – 1)2 + 3 ⇒ f'(x) = 2(x – 1)
Now f'(x) = 0 ⇒ 2 (x – 1) = 0 ⇒ x = 1
Now the critical point is at 1 and end points of [-3, 1]
(i) f(1) = (1 – 1)2 + 3 = 3 ………… (1)
(ii) f (-3) = (-3 – 1)2 + 3 = 16 + 3 = 19 ……………… (2)
Hence, we conclude that the absolute maximum value of f on [-3, 1] is 19 occurring at x = -3 and minimum value of on [-3, 1] is 3 occurring at x = 1.
Question 17.
Find the maximum profit that a company can make, if the profit function is given by p (x) = 41 – 72x – 18x2.
Solution:
Given p(x) = 41 – 72 x – 18x2. …………. (1)
⇒ p'(x) = -72 – 36x ⇒ p'(x) = -36
Now, p'(x) = 0 ⇒ -72 – 36x = 0 ⇒ 36x = -72 ⇒ x = -2
Also p”(2) = -36 < 0
By second derivative test, x = -2 is the point of local maxima of p.
∴ Maximum profit = p(-2)
= 41 – 72(-2) – 18(-2)2 = 41 + 144 – 72 = 113
Hence, the maximum profit that the company can make is 113 units.
Question 18.
Find both the maximum value and the minimum value of 3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3].
Solution:
Given f(x) = 3x4 – 8x3 + 12x2 – 48x + 25
f'(x) = 12x3 – 24x2 + 24x – 48 = 12(x3 – 2x2 + 2x – 4)
= 12[x2 (x – 2) + 2(x – 2)] = 12(x – 2)(x2 + 2)
Now, f'(x) = 0 gives x = 2 or x2 + 2 = 0 for which there are no real roots.
∴ We consider only x = 2 ∈ [0, 3]
Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3]
(i) f(2) = 3(2)4 – 8(2)3 + 12(2)2 – 48(2) + 25
= 48 – 64 + 48 – 96 + 25 = -39 ……… (1)
(ii) f(0) = 3(0)4 – 8(0)3 + 12(0)2 – 48(0) + 25 = 25 ……………. (2)
(iii) f(3) = 3(3)4 – 8(3)3 + 12(3)2 – 48(3) + 25
= 243 – 216 + 108 – 144 + 25 = 16 ……… (3)
Hence, we can conclude that the absolute value of f on [0, 3] is 25 occurring at x = 0 and the absolute minimum value of f at [0, 3] is -39 occurring at x = 2.
Question 19.
At what points in the interval [0, 2π], docs tire function sin 2x attain its maximum value?
Solution:
Given f(x) = sin 2x ⇒ f'(x) = 2 cos 2x
Now, f'(x) = 0 ⇒ 2 cos 2x = 0
⇒ 2x = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\) ⇒ x = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\)
Evaluate the values of f at critical points \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\) and at the end points of the interval [0, 2π].
(i) f\(\left(\frac{\pi}{4}\right)\) = sin\(\left(\frac{\pi}{4}\right)\) = 1
(ii) f\(\left(\frac{3\pi}{4}\right)\) = sin\(\frac{3\pi}{2}\) = -1
(iii) f\(\left(\frac{5\pi}{4}\right)\) = sin \(\frac{5\pi}{2}\) = 1
(iv) f\(\left(\frac{7\pi}{4}\right)\) = sin\(\frac{7\pi}{2}\) = -1
(v) f(0) = sin 0 = 0
(vi) f(2π) = sin 2π = 0
Hence, we conclude that the absolute maximum value of f on [0, 2π] is occurring at x = \(\frac{\pi}{4}\) and x = \(\frac{5\pi}{4}\)
Question 20.
What is the maximum value of the function sin x + cos x?
Solution:
Let f(x) = sin x + cos x ⇒ f'(x) = cos x – sin x
Now, f”(x) = 0 ⇒ cos x – sinx = 0 ⇒ sin x = cos x ⇒ tan x = 1
x = \(\frac{\pi}{4}\), \(\frac{5\pi}{4}\)
Hence, f “(x) = – sin x – cos x = – (sin x + cos x )
Now f”(x) will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive. Also, we know that sin x and cos x both are positive in the first quadrant, then we consider x = \(\frac{\pi}{4}\)
\(f^{\prime \prime}\left(\frac{\pi}{4}\right)=-\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)=-\left(\frac{2}{\sqrt{2}}\right)=-\sqrt{2}\) < 0
By second derivative test, f will be the maximum at x = \(\frac{\pi}{4}\) and the maximum value of f is \(f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Question 21.
Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].
Solution:
Let f(x) = 2x3 – 24x + 107
⇒ f (x) = 6x3 – 24 = 6(x3 – 4)
Now, f'(x) = 0 ⇒ 6(x2 – 4) = 0 ⇒ x2 = 4 ⇒ x = ± 2
Now, we first consider the interval [1, 3].
Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] [and at the end points of the interval [1, 3].
Hence, f(2) = 2(2)3 – 24(2)+107 = 16 – 48 + 107 = 75
f(1) = 2(1)3 – 24(1) + 107 = 2 – 24 + 107 = 85
f(3) = 2(3)3 – 24(3) + 107 = 54 – 72 + 107 = 89
Thus, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.
Next, we consider the interval [-3, -1] and evaluate the value of f at the critical point x = -2
∴ [-3, -1] and at the end points of the interval [1, 3]
f(-3) = 2(-3)3 – 24(-3) + 107 = -54 + 72 + 107 = 125
f(-1) = 2(-1)3 – 24(-1) + 107 = -2 + 24 + 107 = 129
f(-2) = 2(-2)3 – 24(-2) + 107= -16 + 48 + 107 = 139
Hence, the absolute maximum value of f(x) in the interval [-3, -1] is 139 occurring at x = -2
Question 22.
Find the maximum and minimum values of x + sin 2x on [0, 2π].
Solution:
Let f(x) = x ± sin 2x
⇒ f'(x) = 1 + 2 cos 2x
Now f'(x) = 0 = 1 + 2 cos 2x = 0
⇒ cos 2x = 0 ⇒ cos 2x = \(\frac{-1}{2}\) = -cos\(\frac{\pi}{3}\) = cos (π – \(\frac{\pi}{3}\)) = cos \(\frac{2\pi}{3}\)
⇒ 2x = 2nπ ± \(\frac{2\pi}{3}\) [[n ∈ Z] ⇒ x = nπ ± \(\frac{\pi}{3}\) [n ∈ Z]
⇒ x = \(\frac{\pi}{3}\), \(\frac{2\pi}{3}\), \(\frac{4\pi}{3}\), \(\frac{5\pi}{3}\) ∈ [0, 2π]
(v) f(0) = 0 + sin0 = 0
(vi) f(2π) = 2π + sin 4π = 2π + 0 = 2π
Hence, we conclude that the absolute maximum value of f(x) in the interval [0, 2π] is 2π occurring at x = 2π and the absolute minimum value of f(x) in the interval [0, 2π] is 0 occurring at x = 0.
Question 23.
Find two numbers whose sum is 24 and whose product is as large as possible.
Solution:
Let a number be x .
Then, the other number be (24 – x).
Let P(x) denote the product of the two numbers.
Thus, we have: P(x) = x (24 – x) = 24x – x2
∴ P'(x) = 24 – 2x ⇒ P'(x) = -2
Now, P'(x) = 0 ⇒ 24 – 2x = 0 ⇒ 24 = 2x ⇒ x = 12
Also, P'(12) = -2 < 0
By second derivative test, x = 12 is the point of local maxima of P.
Hence, the product of the numbers is the maximum when the numbers are 12 and (24 – 12) = 12.
III.
Question 1.
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.
Solution:
The two numbers are x and y such that x + y = 60 ⇒ y = 60 – x …………. (1)
Let f(x) = xy3 = f(x) = x(60 – x)3 ………………. (1)
⇒ f’(x) = (60 – x)3 – 3x(60 – x)2 = (60 – x)2[60 – x – 3x] = (60 – x)2(60 – 4x)
⇒ f”(x) = -2(6o – x)(6o – 4x) – 4(6o – x)2 = -2(60 – x)[60 – 4x + 2(60 – x)]
= -2(60 – x)(180 – 6x) = -12(60 – x)(30 – x)
Now, f'(x) = 0 ⇒ x = 60 or x = 15
Now, f”(x) = -2 (60 – x) (60 – 4x) – 4(60 – x)2
= -2(60 – x)[60 – 4x + 2(60 – x)]
= -12(60 – x)(30 – x)
When x = 15, f”(x) = -12 (60 – 15) (30 – 15)
= -12 × 45 × 15 < 0
When x = 60, f”(x) = 0
∴ By the second derivative test, x = 15 is a point of local maxima of f.
Thus, function xy3 is maximum when x = 15 and
y = 60 – 15 = 45
Hence, the required numbers are 15 and 45.
Question 2.
Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum.
Solution:
Let a number be x. Then, the other number is y = (35 – x).
Let P(x) = x2y5 – Then we have, P(x) = x2 (35 – x)5
P'(x) = 2x(35 – x)5 + x25(35 – x)4 (-1) = 2x (35 – x)5 – 5x2 (35 – x)4
= x(35 – x)4[2(35 – x) – 5x]
= x(35 – x)4 (70 – 7x) = 7x (35 – x)4(10 – x)
P”(x) = 7(35 – x)4(10 – x) + 7x[-(35 – x)4 – 4(35 – x)3(10 – x)]
= 7(35 – x)4(10 – x) – 7x(35 – x)4 – 28x(35 – x)3 (10 – x)
= 7(35 – x)3[(35 – x)(10 – x) – x(35 – x) – 4x(10 – x)]
= 7(35 – x)3[350 – 45x + x2 – 35x + x2 – 40x + 4x2]
= 7(35 – x)3(6x2 – 120x + 350)
Now, P'(x) = 0 ⇒ x = 0, x = 35, x = 10
When, x = 35 then, P'(x) = P (x) = 0 ⇒ y = 35 – 35 = 0
This will make the product x2y5 equal to 0.
When x = 0 then y = 35 – 0 = 35.This will make the product x2y5 equal to 0.
∴ x = 0 and x = 35 cannot be the possible values of x.
When x = 10, we have
p”(x) = 7 (35 – 10)3 (6 × 100 – 120 × 10 + 350)
= 7 (25)3 (-250) < 0
∴ By second derivative test, P(x) will be the maximum when x = 10 and y = 35 – 10 = 25
Hence, the required numbers are 10 and 25.
Question 3.
Kind two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Solution:
Let a number be x . Then, the other number be (16 – x).
Let the sum of the cubes of these numbers be denoted by S(x).
Then, S(x) = x3 + (16 – x)3
∴ S'(x) = 3x2 + 3(16 – x)2(-1)
= 3x2 – 3(16 – x)2
⇒ S”(x) = 6x + 6(16 – x)
Now, S'(x) = 0 ⇒ 3x2 – 3(16 – x)2 = 0
⇒ x2 – (16 – x)2 = 0
⇒ x2 – 256 – x2 + 32x = 0
⇒ x = \(\frac{256}{32}\) ⇒ x = 8
Also, S”(8) = 6(8) + 6(16 – 8) = 48 + 48 = 96 > 0
∴ By second derivative test, x=8 is the point of local minima of S.
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 – 8 = 8.
Question 4.
A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
Solution:
Let the side of the square to be cut off be x cm.
Then, the length and the breadth of the box will be( 18 – 2x)cm each and the height of the box be x cm.
∴ Volume V (x) of the box is given by,V(x) = lbh = x(18 – 2x)2.
Hence, V'(x) = 1 (18 – 2x )2 + 2x(18 – 2x)(-2)
⇒ V'(x) = (18 – 2x )2 – 4x (18 – 2x )
= (18 – 2x)[18 – 2x – 4x]
= (18 – 2x)(18 – 6x) = 6x2(9 – x)(3 – x) = 12(9 – x)(3 – x)
V”(x) = 12[-(9 – x) – (3 – x)
= -12[9 – x + 3 – x]
= 12(2x – 12) = 24(x – 6)
Now, V'(x) = 0 ⇒ x = 9, x = 3
If, x = 9 then the length and the breadth will become 0.
∴ x ≠ 9, so x = 3
Now, V”(x) = -24(6 – 3) = -72 < 0
By second derivative test, x = 3 is the point of local maxima of V.
Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.
Question 5.
A rectangular sheet of tin 45 cm by 24 cm is to he made into a box without top, by culling off square from each cornet and holding up the flaps. What should he the side of the square to he cut off so that the volume of the box is maximum ?
Solution:
Let the side of the square to be cut off be x cm.
Then, the height of the box is x cm,
the length is (45 – 2x)cm and the breadth (24 – 2x) cm.
∴ Therefore, the volume V(x) of the box is given by,
V(x) = x(45 – 2x)(24 – 2x)
= x(1080 – 90x – 48x + 4x2)
= 4x3 – 138x2 + 1080x
⇒ V'(x) = 12x2 – 276x + 1080
= 12(x2 – 23x + 90) = 12(x – 18) (x – 5)
= 12(x – 18)(x – 5)
V”(x) = = 12(2x – 23) = 24x – 276
Now, V'(x) = 0 ⇒ x = 18, x = 5
It is not possible to cut off a square of side 18 cm from each comer of the rectangular sheet. Thus, x cannot be equal to 18.
∴ x = 5
Then, V”(5) = 12[2 (5) – 23] = 12 (10 – 23) = 12(-13) = -156 < 0
By second derivative test, x = 5 is the point of local maxima.
Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm.
Question 6.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Solution:
Let a rectangle of length 1 and breadth b be inscribed in the given circle of radius a .
Then, the diagonal passes through the centre and is of length 2a cm
Now, by applying the Pythagoras theorem, we have:
By the second derivative test, when l = \(\sqrt{2}\)a, then the area of the rectangle is the maximum.
∴ l = b = \(\sqrt{2}\)a the rectangle is a square.
Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.
Question 7.
Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
Solution:
Let r and h be the radius and height of the cylinder respectively.
Then, the surface area (S) of the cylinder is given by, S = 2πr2 + 2πrh
∴ h = \(\frac{S-2 \pi r^2}{2 \pi r}=\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\)
Let V be the volume of the cylinder
V = πr2h = πr2 = \(\left[\frac{\mathrm{S}}{2 \pi}\left(\frac{1}{\mathrm{r}}\right)-\mathrm{r}\right]\) = \(\frac{\mathrm{Sr}}{2}\) -πr3
⇒ \(\frac{\mathrm{dV}}{\mathrm{dr}}=\frac{\mathrm{S}}{2}\) – 3πr2 ⇒ \(\frac{d^2 V}{d r^2}\) = -6πr
Now, \(\frac{\mathrm{dV}}{\mathrm{dr}}\) = 0 ⇒ \(\frac{\mathrm{S}}{2}\) – 3πr2 = 0
⇒ \(\frac{\mathrm{S}}{2}\) = 3πr2
⇒ r2 = \(\frac{\mathrm{S}}{6 \pi}\)
When r2 = \(\frac{\mathrm{S}}{6 \pi}\)
Then \(\frac{d^2 V}{d r^2}=-6 \pi\left(\sqrt{\frac{S}{6 \pi}}\right)<0\)
By second derivative test, the volume is the maximum when r2 = \(\frac{\mathrm{S}}{6 \pi}\)
Now, when r2 = \(\frac{\mathrm{S}}{6 \pi}\) Then, h = \(\frac{6 \pi \mathrm{r}^2}{2 \pi}\left(\frac{1}{\mathrm{r}}\right)\) – r = 3r – r = 2r
Hence, the volume is the maximum vyhen the height is twice the radius i.e., when the height is equal to the diameter.
Question 8.
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
Solution:
Let r and h be the radius and height of the cylinder respectively.
Then, volume V of the cylinder is given by, V = πr2 = 100 ⇒ h = \(\frac{100}{\pi \mathrm{r}^2}\)
Surface area is given by: S = 2πr2 + 2πrh = 2πr2 + \(\frac{200}{\mathrm{r}}\)
⇒ \(\frac{\mathrm{dS}}{\mathrm{~d} \mathrm{r}}\) = 4πr – \(\frac{200}{r^2}\) ⇒ \(\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{dr}^2}\) = 4πr + \(\frac{400}{r^2}\)
Now, \(\frac{\mathrm{dS}}{\mathrm{dr}}\)= 0 ⇒ 4πr – \(\frac{200}{r^2}\) = 0
⇒ 4πr = \(\frac{200}{r^2}\)
⇒ r3 = \(\frac{200}{4 \pi}=\frac{50}{\pi}\)
⇒ r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\)
Now, it is observed that when
r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) Then, \(\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{dr}^2}\) > 0
∴ By second derivative test, the surface area is the minimum when r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm
when r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\)
h = \(\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{2}{3}}}=\frac{2 \times 50}{(\pi)(50)^{\frac{2}{3}} \cdot \pi^{\frac{2}{3}}}=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm
Hence, the required dimensions of the can which has the minimum surface area is given by radius \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm and height \(2\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm.
Question 9.
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Solution:
Let a piece of length l be cut from the given wire to make a square.
Then, the other piece of wire to be made into a circle is of length (28 – l).
Now, side of square is l/4.
Let r be the radius of the circle.
Then, 2πr = 28 – l ⇒ r = \(\frac{1}{2 \pi}\)(28 – l)
The combined areas of the square and the circle A, is given by,
A = (side of the square)2 + πr2
∴ By second derivative test, the area (A) is the minimum when l = \(\frac{112}{\pi+4}\) cm.
Hence, the combined area is the minimum when the length of the wire in making the square is l = \(\frac{112}{\pi+4}\) cm while the length of the wire in making the circle is \(\left(28-\frac{112}{\pi+4}\right)=\frac{28 \pi}{\pi+4}\) cm.
Question 10.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.
Solution:
Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.
Let V be the volume of the cone.Then V = \(\frac{1}{3}\)πr2 h
Height of the cone is given by, h = R + AB = R + \(\sqrt{R^2-r^2}\) [ABC is a right angle]
⇒ 2R = \(\frac{3 r^2-2 R^2}{\sqrt{R^2-r^2}}\)
⇒ 2R = \(\sqrt{R^2-r^2}\) = 3r2 – 2R2
⇒ 4R2(R2 – r2) = (3r2 – 2R2)2
⇒ 4R4 – 4R2r2 = 9r4 + 4R4 – 12r2R2
⇒ 9r4 = 8R2r2 ⇒ r2 = \(\frac{8}{9}\)R2
When, r2 = \(\frac{8}{9}\) R2. Then \(\frac{d^2 \mathrm{~V}}{\mathrm{dr}^2}\) < 0
By second derivative test, the volume of the cone is the maximum, when r2 = \(\frac{8}{9}\) R2
When, r2 = \(\frac{8}{9}\) R2.
Then, h = R + \(\sqrt{R^2-\frac{8}{9} R^2}=R+\sqrt{\frac{1}{9} R^2}=R+\frac{R}{3}=\frac{4}{3} R\)
∴ V = \(\frac{1}{3} \pi\left(\frac{8}{9} R^2\right)\left(\frac{4}{3} R\right)=\frac{8}{27}\left(\frac{4}{3} \pi R^3\right)=\frac{8}{27} \times(\text { Volume of sphere })\)
Hence, the volume of the largest cone that can be inscribed in the sphere is 8/27 the volume of the sphere.
Question 11.
Show that the right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt{2}\) time the radius of the base.
Solution:
Let r and h be the radius and height of the cone, respectively.
Then, the volume (V) of the cone is given by, V = \(\frac{1}{3}\)πr2 h ⇒ h = \(\frac{3 \mathrm{~V}}{\pi \mathrm{r}^2}\)
The surface area (S) of the cone is given by, S = πrl,
where l is the slant height
Thus, it can be easily verified that when r6 = \(\frac{9 V^2}{2 \pi^2}\), ⇒ \(\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{~d} \mathrm{r}^2}\) > 0
∴ By second derivative test, the surface area of the cone is the least when r6 = \(\frac{9 V^2}{2 \pi^2}\)
When r6 = \(\frac{9 V^2}{2 \pi^2}\)
Then, h = \(\frac{3 \mathrm{~V}}{\pi \mathrm{r}^2}=\frac{3 \mathrm{~V}}{\pi \mathrm{r}^2}\left(\frac{2 \pi^2 \mathrm{r}^6}{9}\right)^{\frac{1}{2}}=\frac{3}{\pi \mathrm{r}^2} \cdot \frac{\sqrt{2} \pi \mathrm{r}^3}{3}=\sqrt{2} \mathrm{r}\)
Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to \(\sqrt{2}\) times the radius of the base.
Question 12.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan-1\(\sqrt{2}\)
Solution:
Let θ be the semi-vertical angle of the cone.
It is clear that θ ∈ [o, \(\frac{\pi}{2}\)]
Let r, h and l be the radius, height, and the slant height of the cone respectively.
The slant height of the cone is given as constant.
Now, r = l sin θ and h = l cos θ
The volume V of the cone is given by, V = \(\frac{1}{3}\) πr2h
= \(\frac{1}{3}\)π(l sinθ)2 (l cos θ) = \(\frac{1}{3}\) πl3 sin2θ . cos θ
= \(\frac{\pi l^3}{3}\)(sin2θ . cosθ)
∴ \(\frac{\mathrm{dV}}{\mathrm{~d} \theta}\) = \(\frac{\pi l^3}{3}\) (sin2θ(-sin θ) + cosθ2sinθ.cosθ)
= \(\frac{\pi l^3}{3}\)(-sin3θ + 2sinθcos2θ)
\(\frac{d^2 V}{d \theta^2}\) = \(\frac{\pi l^3}{3}\)[-3sin2θ cosθ + 2(sin θ . 2 cos θ(-sin θ)) + cos2θ(cos θ)]
= \(\frac{\pi l^3}{3}\) [-3sin2θcosθ – 4sin2θcosθ + 2cos3θ]
= \(\frac{\pi l^3}{3}\) [-7sin2θcosθ + 2cos3θ]
Now, \(\frac{\mathrm{dV}}{\mathrm{~d} \theta}\) = 0
⇒ \(\frac{\pi l^3}{3}\) [-sin3θ + 2sinθcos2θ] = 0
⇒ sin3θ = 2sinθcos2θ
⇒ tan2 θ = \(\sqrt{2}\) since, sin θ ≠ 0
⇒ tan θ = \(\sqrt{2}\) ⇒ θ = tan-1\(\sqrt{2}\)
Now, when θ = tan-1\(\sqrt{2}\), then tan2θ = 2 or
⇒ sin2θ = 2 cos2θ
Then \(\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} \theta^2}=\frac{\pi l^3}{3}\) [2 cos2θ – 14 cos3θ]
= 4πl3 cos3θ < 0 for θ ∈ [o, \(\frac{\pi}{2}\)]
By second derivative test, the volume V is the maximum when θ = tan-1\(\sqrt{2}\)
Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is tan-1\(\sqrt{2}\)
Question 13.
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin-1\(\left(\frac{1}{3}\right)\)
Solution:
Let the height, radius and slant height of th e cone be h, r and l respectively, whose semi-vertical angle is a.
Surface area of cone, S = πrl + πr2
= πr\(\) + πr2
Let V2 = M = \(\frac{1}{9}\)S (Sr3 – 2πr4)
∴ M'(r) = \(\frac{1}{9}\)S (2Sr – 8πr3)
Now, M'(r) = 0 ⇒ \(\frac{1}{9}\)S (2Sr – 8πr3) = 0
= \(\frac{1}{9}\)2Sr (S – 4πr2) = 0
⇒ r = 0 or r2 = \(\frac{\mathrm{S}}{4 \pi}\) ⇒ r = 0 or r = \(\sqrt{\frac{\mathrm{S}}{4 \pi}}\)
[r ≠ 0, as r is the radius of the cone, So r = 0 is not possible]
Now, M”(r) = \(\frac{1}{9}\)S (2S – 24πr2)
For r = \(\sqrt{\frac{\mathrm{S}}{4 \pi}}\)
M”\(\left(\sqrt{\frac{\mathrm{S}}{4 \pi}}\right)\) = \(\frac{1}{9} S\left(2 S-24 \pi\left(\frac{S}{4 \pi}\right)\right)\)
= \(\frac{4}{9}\) S2 < 0
Here, M” \(\left(\sqrt{\frac{\mathrm{S}}{4 \pi}}\right)\) < 0,
∴ r = \(\sqrt{\frac{\mathrm{S}}{4 \pi}}\) is the point of local maxima.
At this maximum point, radius of the cone r = \(\sqrt{\frac{\mathrm{S}}{4 \pi}}\)
⇒ r2 = \(\frac{\mathrm{S}}{4 \pi}\) ⇒ 4πr2 = S
⇒ 4πr2 = πrl + πr2 ⇒ 3r = l
⇒ \(\frac{\mathrm{r}}{l}=\frac{1}{3}\)
Here, sin α = \(\frac{\mathrm{r}}{l}=\frac{1}{3}\)
⇒ α = sin-1\(\left(\frac{1}{3}\right)\)
