GSEB Class 12 Maths Solutions Chapter 6 Application of Derivatives Exercise 6.5

Question 1. Find the maximum and minimum values, if any, of the following functions given by
(i) \( f(x) = (2x - 1)^2 + 3 \)
(ii) \( f(x) = 9x^2 + 12x + 2 \)
(iii) \( f(x) = - (x-1)^2 + 10 \)
(iv) \( g(x) = x^3 + 1 \)
Answer:
(i) For \( f(x) = (2x - 1)^2 + 3 \), the term \( (2x - 1)^2 \) is always greater than or equal to zero. Its minimum value is 0, which happens when \( 2x - 1 = 0 \), so \( x = \frac{1}{2} \). Therefore, the minimum value of \( f(x) \) is \( 0 + 3 = 3 \). Since \( (2x - 1)^2 \) can grow infinitely large, there is no finite maximum value for this function.
In simple words: The smallest this function can be is 3. It can never reach a biggest value.
(ii) For \( f(x) = 9x^2 + 12x + 2 \), we can rewrite it by completing the square: \( f(x) = (3x)^2 + 2(3x)(2) + 2^2 - 2^2 + 2 = (3x + 2)^2 - 4 + 2 = (3x + 2)^2 - 2 \). The term \( (3x + 2)^2 \) is always greater than or equal to zero. Its minimum value is 0, when \( 3x + 2 = 0 \), so \( x = -\frac{2}{3} \). Therefore, the minimum value of \( f(x) \) is \( 0 - 2 = -2 \). Similar to part (i), \( (3x + 2)^2 \) can grow infinitely large, so there is no finite maximum value for this function.
In simple words: This function is at least -2, but it does not have an upper limit.
(iii) For \( f(x) = - (x - 1)^2 + 10 \), the term \( (x - 1)^2 \) is always greater than or equal to zero. Thus, \( - (x - 1)^2 \) is always less than or equal to zero. Its maximum value is 0, occurring when \( x - 1 = 0 \), so \( x = 1 \). Therefore, the maximum value of \( f(x) \) is \( 0 + 10 = 10 \). Since \( (x - 1)^2 \) can grow infinitely large in the positive direction, \( - (x - 1)^2 \) can grow infinitely large in the negative direction. This means there is no finite minimum value for this function.
In simple words: The biggest this function can be is 10. It keeps getting smaller without limit.
(iv) For \( g(x) = x^3 + 1 \), as \( x \) approaches positive infinity, \( x^3 \) approaches positive infinity, so \( g(x) \) also approaches positive infinity. As \( x \) approaches negative infinity, \( x^3 \) approaches negative infinity, so \( g(x) \) also approaches negative infinity. The derivative \( g'(x) = 3x^2 \). Since \( g'(x) \ge 0 \) for all \( x \), the function \( g(x) \) is always increasing. Because it continuously increases from negative infinity to positive infinity, it does not have any local or absolute maximum or minimum values.
In simple words: This function always goes up, from negative infinity to positive infinity, so it has no highest or lowest point.

Exam Tip: For quadratic functions like \( ax^2+bx+c \), if \( a>0 \), there is a minimum value; if \( a<0 \), there is a maximum value. For cubic functions like \( x^3+c \), there are generally no global max/min unless an interval is specified.

Question 2. Find the maximum and minimum values, if any, of the following functions given, by
(i) \( f(x) = |x + 2| - 1 \)
(ii) \( g(x) = -|x + 1| + 3 \)
(iii) \( h(x) = \sin (2x) + 5 \)
(iv) \( f(x) = \sin (4x) + 3 \)
(v) \( h(x) = x + 1, x \in (-1,1) \)
Answer:
(i) For \( f(x) = |x + 2| - 1 \), the term \( |x + 2| \) is always greater than or equal to 0 for any real \( x \). The minimum value of \( |x + 2| \) is 0, which occurs when \( x = -2 \). Therefore, the minimum value of \( f(x) \) is \( 0 - 1 = -1 \). This function does not have any maximum value because \( |x + 2| \) can be arbitrarily large.
In simple words: The function's lowest value is -1, but it can grow infinitely large.
(ii) For \( g(x) = -|x + 1| + 3 \), we know that \( |x + 1| \ge 0 \) for all real numbers \( x \). This implies that \( -|x + 1| \le 0 \) for all \( x \). Adding 3 to both sides, we get \( -|x + 1| + 3 \le 3 \), so \( g(x) \le 3 \). Thus, the maximum value of \( g(x) \) is 3, achieved when \( x = -1 \). There is no minimum value for this function because \( -|x + 1| \) can be arbitrarily small (large negative).
In simple words: The function's highest value is 3, but it can get infinitely small.
(iii) For \( h(x) = \sin (2x) + 5 \), the sine function, \( \sin (2x) \), has a maximum value of 1 and a minimum value of -1. Therefore, the maximum value of \( h(x) \) is \( 1 + 5 = 6 \). And the minimum value of \( h(x) \) is \( -1 + 5 = 4 \).
In simple words: The highest the function can be is 6, and the lowest it can be is 4.
(iv) For \( f(x) = \sin (4x) + 3 \), the maximum value of \( \sin (4x) \) is 1. Therefore, the maximum value of \( f(x) = \sin (4x) + 3 \) is \( 1 + 3 = 4 \). The minimum value of \( \sin (4x) \) is -1. Thus, the minimum value of \( f(x) = \sin (4x) + 3 \) is \( -1 + 3 = 2 \).
In simple words: The greatest value is 4, and the smallest value is 2.
(v) For \( h(x) = x + 1 \) for \( x \in (-1, 1) \), the function is strictly increasing on this open interval. As \( x \) approaches 1, \( h(x) \) approaches \( 1 + 1 = 2 \). As \( x \) approaches -1, \( h(x) \) approaches \( -1 + 1 = 0 \). Since the interval is open, the function does not actually reach these values. Therefore, there is no absolute maximum or absolute minimum value within the specified open interval. The greatest value it approaches is 2, and the least value it approaches is 0.
In simple words: This function keeps getting bigger but never quite reaches 2. It keeps getting smaller but never quite reaches 0. So, it has no exact biggest or smallest point.

Exam Tip: For functions involving absolute values or trigonometric terms, recall their inherent range. For open intervals, maximum and minimum values may not be attained but approached as limits.

Question 3. Find the local maxima or local minima, if any, of the following functions. Find also the local maximum and local minimum values, as the case may be:
(i) \( f(x) = x^2 \)
(ii) \( g(x) = x^3 - 3x \)
(iii) \( h(x) = \sin x + \cos x, 0 < x < \frac { \pi }{ 2 } \)
(iv) \( f(x) = \sin x - \cos x, 0 < x < 2\pi \)
(v) \( f(x) = x^3 - 6x^2 + 9x + 15 \)
(vi) \( g(x)= \frac { x }{ 2 } + \frac {2}{x}, x > 0 \)
(vii) \( g(x) = \frac{1}{x^{2}+2} \)
(viii) \( f(x) = x\sqrt{1-x}, x > 0 \)
Answer:
(i) Let \( f(x) = x^2 \). The first derivative is \( f'(x) = 2x \). Setting \( f'(x) = 0 \) gives \( 2x = 0 \), which means \( x = 0 \). At \( x = 0 \): When \( x \) is slightly less than 0, \( f'(x) \) is negative. When \( x \) is slightly greater than 0, \( f'(x) \) is positive. Thus, \( f'(x) \) changes its sign from negative to positive as \( x \) passes through 0. This indicates that \( f(x) \) has a local minimum at \( x = 0 \). The local minimum value is \( f(0) = 0^2 = 0 \).
In simple words: The function has its lowest point at \( x=0 \), where its value is 0.
(ii) Let \( g(x) = x^3 - 3x \). The derivative is \( g'(x) = 3x^2 - 3 \), which factors as \( 3(x - 1)(x + 1) \). Setting \( g'(x) = 0 \) yields \( x = 1 \) or \( x = -1 \). At \( x = 1 \): When \( x \) is slightly less than 1, \( g'(x) \) is negative. When \( x \) is slightly greater than 1, \( g'(x) \) is positive. Hence, \( g'(x) \) changes its sign from negative to positive as \( x \) moves past 1. This means \( x = 1 \) is a point of local minima. The local minimum value is \( g(1) = 1^3 - 3(1) = 1 - 3 = -2 \). At \( x = -1 \): When \( x \) is slightly less than -1, \( g'(x) \) is positive. When \( x \) is slightly greater than -1, \( g'(x) \) is negative. Consequently, \( g'(x) \) changes its sign from positive to negative as \( x \) passes -1. This indicates that \( x = -1 \) is a point of local maxima. The local maximum value is \( g(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2 \).
In simple words: This function has a low point at \( x=1 \) (value -2) and a high point at \( x=-1 \) (value 2).
(iii) Let \( h(x) = \sin x + \cos x \) for \( 0 < x < \frac{\pi}{2} \). The derivative is \( h'(x) = \cos x - \sin x \). We can rewrite this as \( \cos x (1 - \tan x) \). Setting \( h'(x) = 0 \) gives \( \cos x - \sin x = 0 \), which implies \( \tan x = 1 \). In the interval \( (0, \frac{\pi}{2}) \), this occurs at \( x = \frac{\pi}{4} \). At \( x = \frac{\pi}{4} \): When \( x \) is slightly less than \( \frac{\pi}{4} \), \( \cos x \) is positive and \( (1 - \tan x) \) is positive, making \( h'(x) \) positive. When \( x \) is slightly greater than \( \frac{\pi}{4} \), \( \cos x \) is positive, but \( \tan x > 1 \), so \( (1 - \tan x) \) becomes negative. This makes \( h'(x) \) negative. Therefore, \( h'(x) \) changes its sign from positive to negative as \( x \) passes through \( \frac{\pi}{4} \). This signifies that there is a local maximum at \( x = \frac{\pi}{4} \). The local maximum value is \( h(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \).
In simple words: The function reaches its highest point in this range at \( x = \frac{\pi}{4} \), and that value is \( \sqrt{2} \).
(iv) Let \( f(x) = \sin x - \cos x \) for \( x \in (0, 2\pi) \). The derivative is \( f'(x) = \cos x + \sin x \). Setting \( f'(x) = 0 \) means \( \cos x + \sin x = 0 \), or \( \tan x = -1 \). In the interval \( (0, 2\pi) \), the critical points are \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \). At \( x = \frac{3\pi}{4} \): When \( x \) is slightly less than \( \frac{3\pi}{4} \), \( f'(x) = \cos x + \sin x \) is positive. When \( x \) is slightly greater than \( \frac{3\pi}{4} \), \( f'(x) \) is negative. Therefore, \( f'(x) \) changes its sign from positive to negative, indicating a local maximum at \( x = \frac{3\pi}{4} \). The local maximum value is \( f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{4}) - \cos(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) = \frac{2}{\sqrt{2}} = \sqrt{2} \). At \( x = \frac{7\pi}{4} \): When \( x \) is slightly less than \( \frac{7\pi}{4} \), \( f'(x) = \cos x + \sin x \) is negative. When \( x \) is slightly greater than \( \frac{7\pi}{4} \) (considering the periodic nature just outside the interval), \( f'(x) \) is positive. Therefore, \( f'(x) \) changes its sign from negative to positive, indicating a local minimum at \( x = \frac{7\pi}{4} \). The local minimum value is \( f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{4}) - \cos(\frac{7\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2} \).
In simple words: This function has a local high point of \( \sqrt{2} \) at \( x = \frac{3\pi}{4} \) and a local low point of \( -\sqrt{2} \) at \( x = \frac{7\pi}{4} \).
(v) Let \( f(x) = x^3 - 6x^2 + 9x + 15 \). The derivative is \( f'(x) = 3x^2 - 12x + 9 \), which can be factored as \( 3(x^2 - 4x + 3) = 3(x - 1)(x - 3) \). Setting \( f'(x) = 0 \) gives \( x = 1 \) or \( x = 3 \). At \( x = 1 \): When \( x \) is slightly less than 1, \( f'(x) \) is positive \( ((-)(-) = +) \). When \( x \) is slightly greater than 1, \( f'(x) \) is negative \( ((+)(-) = -) \). Thus, \( f'(x) \) changes its sign from positive to negative as \( x \) increases through 1. This means \( f(x) \) has a local maximum at \( x = 1 \). The local maximum value is \( f(1) = 1^3 - 6(1)^2 + 9(1) + 15 = 1 - 6 + 9 + 15 = 19 \). At \( x = 3 \): When \( x \) is slightly less than 3, \( f'(x) \) is negative \( ((+)(-) = -) \). When \( x \) is slightly greater than 3, \( f'(x) \) is positive \( ((+)(+) = +) \). So \( f'(x) \) changes its sign from negative to positive as \( x \) increases through 3. This indicates \( f(x) \) has a local minimum at \( x = 3 \). The local minimum value is \( f(3) = (3)^3 - 6(3)^2 + 9(3) + 15 = 27 - 54 + 27 + 15 = 15 \).
In simple words: The function has a local peak of 19 at \( x=1 \) and a local valley of 15 at \( x=3 \).
(vi) Let \( g(x) = \frac{x}{2} + \frac{2}{x} \) for \( x > 0 \). The derivative is \( g'(x) = \frac{1}{2} - \frac{2}{x^2} \), which simplifies to \( \frac{x^2 - 4}{2x^2} \). Setting \( g'(x) = 0 \) means \( x^2 - 4 = 0 \), so \( x^2 = 4 \), which gives \( x = \pm 2 \). Since the domain is \( x > 0 \), we reject \( x = -2 \), leaving \( x = 2 \). For \( x \) slightly less than 2, \( g'(x) \) is negative. For \( x \) slightly greater than 2, \( g'(x) \) is positive. Therefore, \( g'(x) \) changes its sign from negative to positive as \( x \) passes through 2. This implies \( g(x) \) has a local minimum at \( x = 2 \). The local minimum value is \( g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2 \).
In simple words: For positive \( x \), the function's lowest point is at \( x=2 \), where its value is 2.
(vii) Let \( g(x) = \frac{1}{x^2+2} \). The derivative is \( g'(x) = - \frac{2x}{(x^2+2)^2} \). Setting \( g'(x) = 0 \) gives \( x = 0 \). At \( x = 0 \): When \( x \) is slightly less than 0, \( -2x \) is positive, so \( g'(x) \) is positive. When \( x \) is slightly greater than 0, \( -2x \) is negative, so \( g'(x) \) is negative. Therefore, \( g'(x) \) changes its sign from positive to negative as \( x \) increases through 0. This indicates that \( g(x) \) has a local maximum at \( x = 0 \). The local maximum value is \( g(0) = \frac{1}{0^2+2} = \frac{1}{2} \).
In simple words: The function reaches its highest local point at \( x=0 \), with a value of \( \frac{1}{2} \).
(viii) Let \( f(x) = x\sqrt{1-x} \) for \( x \in (0, 1] \). The derivative is \( f'(x) = 1 \cdot \sqrt{1-x} + x \cdot \frac{1}{2\sqrt{1-x}}(-1) \). This simplifies to \( f'(x) = \frac{2(1-x) - x}{2\sqrt{1-x}} = \frac{2 - 3x}{2\sqrt{1-x}} \). Setting \( f'(x) = 0 \) gives \( 2 - 3x = 0 \), so \( x = \frac{2}{3} \). At \( x = \frac{2}{3} \): When \( x \) is slightly less than \( \frac{2}{3} \), \( (2 - 3x) \) is positive, so \( f'(x) \) is positive. When \( x \) is slightly greater than \( \frac{2}{3} \), \( (2 - 3x) \) is negative, so \( f'(x) \) is negative. Therefore, \( f'(x) \) changes its sign from positive to negative as \( x \) increases through \( \frac{2}{3} \). This indicates that \( f(x) \) has a local maximum at \( x = \frac{2}{3} \). The local maximum value is \( f(\frac{2}{3}) = \frac{2}{3}\sqrt{1-\frac{2}{3}} = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3\sqrt{3}} \).
In simple words: The function has a local highest point at \( x = \frac{2}{3} \), with a value of \( \frac{2}{3\sqrt{3}} \).

Exam Tip: Always analyze the sign change of the first derivative around critical points to determine if it's a local maximum (positive to negative) or local minimum (negative to positive). For square root functions, remember to consider the domain carefully.

Question 4. Prove that the following functions do not have maxima or minima :
(i) \( f(x) = e^x \)
(ii) \( g(x) = \log x \)
(iii) \( h(x) = x^3 + x^2 + x + 1 \)
Answer:
(i) For \( f(x) = e^x \), the derivative is \( f'(x) = e^x \). For local maxima or minima, we set \( f'(x) = 0 \), which implies \( e^x = 0 \). However, \( e^x \) is always positive and never equals zero for any finite real value of \( x \). Since there are no critical points where \( f'(x) = 0 \), the function \( f(x) = e^x \) does not possess any local maxima or minima.
In simple words: Since \( e^x \) is always positive and never zero, this function never has a turning point.
(ii) For \( g(x) = \log x \), the derivative is \( g'(x) = \frac{1}{x} \). For local maxima or minima, we would set \( g'(x) = 0 \), which means \( \frac{1}{x} = 0 \). This equation has no solution for any finite real value of \( x \). Additionally, the domain of \( \log x \) is \( x > 0 \), and \( g'(x) \) is never zero or undefined within its domain. Since \( g'(x) \) is always positive for \( x > 0 \), the function \( g(x) \) is strictly increasing. Therefore, \( g(x) = \log x \) does not have any local maxima or minima.
In simple words: The derivative is never zero, and the function always increases, so it has no highest or lowest points.
(iii) Let \( h(x) = x^3 + x^2 + x + 1 \). The derivative is \( h'(x) = 3x^2 + 2x + 1 \). To find critical points, we set \( h'(x) = 0 \), so \( 3x^2 + 2x + 1 = 0 \). Using the quadratic formula, \( x = \frac{-2 \pm \sqrt{2^2 - 4(3)(1)}}{2(3)} = \frac{-2 \pm \sqrt{4 - 12}}{6} = \frac{-2 \pm \sqrt{-8}}{6} \). Since the discriminant is negative, the roots are imaginary. This means \( h'(x) \) is never zero for any real value of \( x \). As \( h'(x) = 3x^2 + 2x + 1 \) is an upward-opening parabola with no real roots, it is always positive. Therefore, \( h(x) \) is strictly increasing and has neither a local maximum nor a local minimum.
In simple words: The derivative of this function is always positive, meaning the function always rises and never has a turning point.

Exam Tip: Functions without real critical points (where \( f'(x)=0 \) or is undefined) or functions that are strictly monotonic (always increasing or always decreasing) across their domain will not have local maxima or minima.

Question 5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
(i) \( f(x) = x^3[-2, 2] \)
(ii) \( f(x) = \sin x + \cos x, x \in [0, \pi] \)
(iii) \( f(x) = 4x - x^2, x \in [-2, \frac { 9 }{ 2 }] \)
(iv) \( f(x) = (x - 1)^2 + 3, x \in [-3, 1] \)
Answer:
(i) Let \( f(x) = x^3 \) in the interval \( [-2, 2] \). The derivative is \( f'(x) = 3x^2 \). Setting \( f'(x) = 0 \) gives \( x = 0 \), so \( f(0) = 0 \). Now we evaluate the function at the critical point and the interval endpoints: \( f(-2) = (-2)^3 = -8 \) \( f(0) = (0)^3 = 0 \) \( f(2) = (2)^3 = 8 \) Comparing these values, the absolute maximum value of \( f(x) \) is 8, which occurs at \( x = 2 \), and the absolute minimum value of \( f(x) \) is -8, which occurs at \( x = -2 \).
In simple words: The highest value for this function in the given range is 8, and the lowest value is -8.
(ii) Let \( f(x) = \sin x + \cos x \) in the interval \( [0, \pi] \). The derivative is \( f'(x) = \cos x - \sin x \). For extreme values, we set \( f'(x) = 0 \), which means \( \cos x - \sin x = 0 \), or \( \cos x = \sin x \). This implies \( \tan x = 1 \). In the interval \( [0, \pi] \), this occurs at \( x = \frac{\pi}{4} \). We find the values of \( f(x) \) at the critical point and the interval endpoints: \( x = 0, \frac{\pi}{4}, \pi \). \( f(0) = \sin 0 + \cos 0 = 0 + 1 = 1 \). \( f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \). \( f(\pi) = \sin \pi + \cos \pi = 0 - 1 = -1 \). Comparing these values, the absolute maximum value of \( f(x) \) is \( \sqrt{2} \), attained at \( x = \frac{\pi}{4} \), and the absolute minimum value is -1, attained at \( x = \pi \).
In simple words: The function's highest value is \( \sqrt{2} \), and its lowest value is -1 within this interval.
(iii) Let \( f(x) = 4x - x^2 \) in the interval \( [-2, \frac{9}{2}] \). The derivative is \( f'(x) = 4 - 2x \). Setting \( f'(x) = 0 \) gives \( 4 - 2x = 0 \), so \( x = 2 \). Now, we evaluate the function at the critical point (\( x=2 \)) and the endpoints (\( x=-2, x=\frac{9}{2} \)): \( f(-2) = 4(-2) - (-2)^2 = -8 - 4 = -12 \). \( f(2) = 4(2) - (2)^2 = 8 - 4 = 4 \). \( f(\frac{9}{2}) = 4(\frac{9}{2}) - (\frac{9}{2})^2 = 18 - \frac{81}{4} = \frac{72 - 81}{4} = -\frac{9}{4} = -2.25 \). Comparing these values, the absolute maximum value of \( f(x) \) is 4, attained at \( x = 2 \), and the absolute minimum value of \( f(x) \) is -12, attained at \( x = -2 \).
In simple words: The maximum value is 4 (at \( x=2 \)), and the minimum value is -12 (at \( x=-2 \)) in the given range.
(iv) Let \( f(x) = (x - 1)^2 + 3 \) in the interval \( [-3, 1] \). The derivative is \( f'(x) = 2(x - 1) \). Setting \( f'(x) = 0 \) gives \( x - 1 = 0 \), so \( x = 1 \). Now, we evaluate the function at the critical point (\( x=1 \)) and the endpoint (\( x=-3 \)): \( f(1) = (1 - 1)^2 + 3 = 0 + 3 = 3 \). \( f(-3) = (-3 - 1)^2 + 3 = (-4)^2 + 3 = 16 + 3 = 19 \). Comparing these values, the absolute maximum value of \( f(x) \) is 19, attained at \( x = -3 \), and the absolute minimum value of \( f(x) \) is 3, attained at \( x = 1 \).
In simple words: The highest value this function reaches is 19, and its lowest value is 3 in the specified interval.

Exam Tip: To find absolute maximum and minimum values on a closed interval, always evaluate the function at critical points within the interval and at the endpoints of the interval. The largest and smallest of these values will be the absolute extrema.

Question 6. Find the maximum profit that a company can make, if the profit function is given by \( p(x) = 41-24x-18x^2 \).
Answer: The profit function is given by \( p(x) = 41 - 24x - 18x^2 \). To find the maximum profit, we first compute the derivative \( p'(x) \). \( p'(x) = -24 - 36x = -12(2 + 3x) \). For local maxima or minima, we set \( p'(x) = 0 \): \( -12(2 + 3x) = 0 \)
\( \implies 2 + 3x = 0 \)
\( \implies 3x = -2 \)
\( \implies x = -\frac{2}{3} \). To verify it's a maximum, we can examine the sign change of \( p'(x) \) around \( x = -\frac{2}{3} \). If \( x < -\frac{2}{3} \), then \( (2+3x) < 0 \), so \( p'(x) = -12(\text{negative}) \) is positive. If \( x > -\frac{2}{3} \), then \( (2+3x) > 0 \), so \( p'(x) = -12(\text{positive}) \) is negative. Since \( p'(x) \) changes from positive to negative, \( p(x) \) has a local maximum at \( x = -\frac{2}{3} \). The maximum profit is \( p(-\frac{2}{3}) = 41 - 24(-\frac{2}{3}) - 18(-\frac{2}{3})^2 \). \( = 41 + 16 - 18(\frac{4}{9}) \) \( = 41 + 16 - 8 \) \( = 49 \). So, the maximum profit a company can make is Rs. 49.
In simple words: To find the highest profit, we first found where the profit changes direction. This happened at \( x = -\frac{2}{3} \). Plugging this value back into the profit formula gave us the maximum profit of Rs. 49.

Exam Tip: For profit maximization problems, find the derivative of the profit function, set it to zero to find critical points, and then use the second derivative test or sign analysis to confirm it's a maximum.

Question 7. Find both the maximum and minimum values of \( 3x^4 - 8x^3 + 12x^2 - 48x + 25 \) on the interval [0, 3].
Answer: Let \( f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25 \) on the interval \( [0, 3] \). First, we find the derivative: \( f'(x) = 12x^3 - 24x^2 + 24x - 48 \) \( = 12(x^3 - 2x^2 + 2x - 4) \) \( = 12[x^2(x - 2) + 2(x - 2)] \) \( = 12(x^2 + 2)(x - 2) \). For local maxima or minima, we set \( f'(x) = 0 \): \( 12(x^2 + 2)(x - 2) = 0 \). Since \( x^2 + 2 \) is always positive, the only real critical point is \( x = 2 \). Now, we evaluate the function at the critical point (\( x=2 \)) and the interval endpoints (\( x=0, x=3 \)): \( f(0) = 3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25 \). \( f(2) = 3(2)^4 - 8(2)^3 + 12(2)^2 - 48(2) + 25 \) \( = 3(16) - 8(8) + 12(4) - 96 + 25 \) \( = 48 - 64 + 48 - 96 + 25 = -39 \). \( f(3) = 3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25 \) \( = 3(81) - 8(27) + 12(9) - 144 + 25 \) \( = 243 - 216 + 108 - 144 + 25 = 16 \). Comparing these values, the maximum value of \( f(x) \) on \( [0, 3] \) is 25 (at \( x=0 \)), and the minimum value is -39 (at \( x=2 \)).
In simple words: We found the function's turning points and checked its values at those points and the ends of the interval. The highest value was 25, and the lowest was -39.

Exam Tip: When finding absolute extrema on a closed interval, remember to consider both critical points within the interval and the function's values at the interval's endpoints.

Question 8. At what points in the interval [0, 2π], does the function \( \sin(2x) \) attains its maximum value?
Answer: Let \( f(x) = \sin(2x) \) in the interval \( [0, 2\pi] \). The derivative is \( f'(x) = 2 \cos(2x) \). For local maxima or minima, we set \( f'(x) = 0 \), which implies \( 2 \cos(2x) = 0 \), so \( \cos(2x) = 0 \). Since \( x \in [0, 2\pi] \), \( 2x \in [0, 4\pi] \). The values for \( 2x \) where \( \cos(2x) = 0 \) are: \( 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} \). Dividing by 2, the critical points for \( x \) are: \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \). Now, we evaluate \( f(x) \) at these critical points and the interval endpoints (\( x=0, x=2\pi \)): \( f(0) = \sin(2 \cdot 0) = \sin 0 = 0 \). \( f(\frac{\pi}{4}) = \sin(2 \cdot \frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1 \). \( f(\frac{3\pi}{4}) = \sin(2 \cdot \frac{3\pi}{4}) = \sin(\frac{3\pi}{2}) = -1 \). \( f(\frac{5\pi}{4}) = \sin(2 \cdot \frac{5\pi}{4}) = \sin(\frac{5\pi}{2}) = 1 \). \( f(\frac{7\pi}{4}) = \sin(2 \cdot \frac{7\pi}{4}) = \sin(\frac{7\pi}{2}) = -1 \). \( f(2\pi) = \sin(2 \cdot 2\pi) = \sin(4\pi) = 0 \). Comparing these values, the maximum value of \( f(x) \) is 1, which is attained at \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).
In simple words: To find where the function \( \sin(2x) \) is highest between 0 and \( 2\pi \), we checked its turning points and the ends of the range. The maximum value is 1, which happens at \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

Exam Tip: When dealing with trigonometric functions over an interval, remember to find all critical points within the given range for both the function and its argument (e.g., \( 2x \) in this case).

Question 9. What is the maximum value of the function \( \sin x + \cos x \)?
Answer: Let \( f(x) = \sin x + \cos x \). To find the maximum value, we can consider the interval \( [0, 2\pi] \). First, we find the derivative: \( f'(x) = \cos x - \sin x \). For local maxima or minima, we set \( f'(x) = 0 \): \( \cos x - \sin x = 0 \)
\( \implies \cos x = \sin x \)
\( \implies \tan x = 1 \). In the interval \( [0, 2\pi] \), the critical points are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \). Now, we evaluate \( f(x) \) at these critical points and the interval endpoints (\( x=0, x=2\pi \)): \( f(0) = \sin 0 + \cos 0 = 0 + 1 = 1 \). \( f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \). \( f(\frac{5\pi}{4}) = \sin(\frac{5\pi}{4}) + \cos(\frac{5\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2} \). \( f(2\pi) = \sin(2\pi) + \cos(2\pi) = 0 + 1 = 1 \). Comparing these values, the maximum value of \( f(x) \) is \( \sqrt{2} \).
In simple words: By checking the function's values at its turning points and the ends of a full cycle, we found its greatest value is \( \sqrt{2} \).

Exam Tip: The function \( A \sin x + B \cos x \) can be expressed in the form \( R \sin(x+\alpha) \), where \( R = \sqrt{A^2+B^2} \). Its maximum value is \( R \) and minimum is \( -R \). For \( \sin x + \cos x \), \( R = \sqrt{1^2+1^2} = \sqrt{2} \), so the maximum is \( \sqrt{2} \).

Question 10. Find the maximum value of \( 2x^3 - 24x +107 \) in the interval [1, 3]. Find the maximum value of the same function in [- 3, -1].
Answer: Let the function be \( f(x) = 2x^3 - 24x + 107 \). The derivative is \( f'(x) = 6x^2 - 24 \). For local maxima or minima, we set \( f'(x) = 0 \): \( 6x^2 - 24 = 0 \)
\( \implies 6x^2 = 24 \)
\( \implies x^2 = 4 \)
\( \implies x = \pm 2 \). **For the interval \( [1, 3] \):** We evaluate \( f(x) \) at the critical point \( x=2 \) (which is in the interval) and the endpoints \( x=1, x=3 \): \( f(1) = 2(1)^3 - 24(1) + 107 = 2 - 24 + 107 = 85 \). \( f(2) = 2(2)^3 - 24(2) + 107 = 16 - 48 + 107 = 75 \). \( f(3) = 2(3)^3 - 24(3) + 107 = 54 - 72 + 107 = 89 \). Comparing these values, the maximum value of \( f(x) \) in \( [1, 3] \) is 89, attained at \( x = 3 \). **For the interval \( [-3, -1] \):** We evaluate \( f(x) \) at the critical point \( x=-2 \) (which is in the interval) and the endpoints \( x=-3, x=-1 \): \( f(-3) = 2(-3)^3 - 24(-3) + 107 = 2(-27) + 72 + 107 = -54 + 72 + 107 = 125 \). \( f(-2) = 2(-2)^3 - 24(-2) + 107 = 2(-8) + 48 + 107 = -16 + 48 + 107 = 139 \). \( f(-1) = 2(-1)^3 - 24(-1) + 107 = 2(-1) + 24 + 107 = -2 + 24 + 107 = 129 \). Comparing these values, the maximum value of \( f(x) \) in \( [-3, -1] \) is 139, attained at \( x = -2 \).
In simple words: We calculated the function's highest value for two different ranges. In the first range [1, 3], the maximum was 89. In the second range [-3, -1], the maximum was 139.

Exam Tip: When asked for maximum/minimum values on a given interval, always check the function's values at any critical points within that interval, as well as at the interval's endpoints.

Question 11. It is given that at \( x = 1 \), the function \( x^4 - 62x^2 + ax + 9 \) attains its maximum value, on the interval [0, 2]. Find the value of a.
Answer: Let the function be \( f(x) = x^4 - 62x^2 + ax + 9 \). The first derivative is \( f'(x) = 4x^3 - 124x + a \). We are given that \( f(x) \) attains its maximum value at \( x = 1 \). For a maximum, the first derivative must be zero at this point. So, \( f'(1) = 0 \): \( 4(1)^3 - 124(1) + a = 0 \) \( 4 - 124 + a = 0 \) \( -120 + a = 0 \)
\( \implies a = 120 \). To confirm this is a maximum, we check the second derivative: \( f''(x) = 12x^2 - 124 \). At \( x = 1 \): \( f''(1) = 12(1)^2 - 124 = 12 - 124 = -112 \). Since \( f''(1) = -112 < 0 \), it confirms that \( f(x) \) indeed has a local maximum at \( x = 1 \). Therefore, the value of \( a \) is 120.
In simple words: Since the function peaks at \( x=1 \), its rate of change (derivative) must be zero there. We used this fact to solve for 'a', finding it to be 120, and then confirmed it was a peak using the second derivative test.

Exam Tip: If a function has an extremum (maximum or minimum) at a point, its first derivative at that point must be zero. The second derivative test can then confirm the nature of that extremum.

Question 12. Find the maximum and minimum values of \( x + \sin(2x) \) on [0, 2π].
Answer: Let \( f(x) = x + \sin(2x) \) on the interval \( [0, 2\pi] \). The first derivative is \( f'(x) = 1 + 2\cos(2x) \). For critical points, we set \( f'(x) = 0 \): \( 1 + 2\cos(2x) = 0 \)
\( \implies \cos(2x) = -\frac{1}{2} \). Since \( x \in [0, 2\pi] \), \( 2x \in [0, 4\pi] \). The values for \( 2x \) where \( \cos(2x) = -\frac{1}{2} \) are: \( 2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3} \). Dividing by 2, the critical points for \( x \) are: \( x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \). Now, we evaluate \( f(x) \) at these critical points and the interval endpoints (\( x=0, x=2\pi \)): \( f(0) = 0 + \sin(0) = 0 \). \( f(\frac{\pi}{3}) = \frac{\pi}{3} + \sin(\frac{2\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \). \( f(\frac{2\pi}{3}) = \frac{2\pi}{3} + \sin(\frac{4\pi}{3}) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \). \( f(\frac{4\pi}{3}) = \frac{4\pi}{3} + \sin(\frac{8\pi}{3}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2} \). \( f(\frac{5\pi}{3}) = \frac{5\pi}{3} + \sin(\frac{10\pi}{3}) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2} \). \( f(2\pi) = 2\pi + \sin(4\pi) = 2\pi \). Approximate values: \( f(0) = 0 \), \( f(\frac{\pi}{3}) \approx 1.91 \), \( f(\frac{2\pi}{3}) \approx 1.23 \), \( f(\frac{4\pi}{3}) \approx 5.05 \), \( f(\frac{5\pi}{3}) \approx 4.37 \), \( f(2\pi) \approx 6.28 \). Comparing all these values, the maximum value of \( f(x) \) is \( 2\pi \) (at \( x=2\pi \)), and the minimum value of \( f(x) \) is 0 (at \( x=0 \)).
In simple words: We found the points where the function's slope is zero and checked the function's value there, as well as at the start and end of the given range. The highest value was \( 2\pi \) and the lowest was 0.

Exam Tip: When evaluating trigonometric functions, remember their periodic nature and the range of values they can take. Always check endpoints and critical points thoroughly.

Question 13. Find two numbers whose sum is 24 and whose product is as large as possible.
Answer: Let the two numbers be \( x \) and \( 24 - x \). Their product, \( P \), is given by \( P(x) = x(24 - x) = 24x - x^2 \). To maximize the product, we find the derivative \( \frac{dP}{dx} \): \( \frac{dP}{dx} = 24 - 2x \). Set \( \frac{dP}{dx} = 0 \) to find critical points: \( 24 - 2x = 0 \)
\( \implies 2x = 24 \)
\( \implies x = 12 \). Now, we find the second derivative \( \frac{d^2P}{dx^2} \): \( \frac{d^2P}{dx^2} = -2 \). Since \( \frac{d^2P}{dx^2} = -2 < 0 \), the product \( P \) is maximum when \( x = 12 \). The first number is \( x = 12 \). The second number is \( 24 - x = 24 - 12 = 12 \). Therefore, the two numbers are 12 and 12.
In simple words: We defined the product using one number, found the derivative, and set it to zero. This showed that the largest product happens when both numbers are 12.

Exam Tip: For optimization problems, define the quantity to be optimized as a function of one variable. Use calculus (first and second derivatives) to find the extremum and confirm its nature.

Question 14. Find two positive numbers x and y such that \( x + y = 60 \) and \( xy³ \) is maximum.
Answer: Let the two numbers be \( x \) and \( y \). We are given that their sum is 60, so \( x + y = 60 \). This means we can express \( y \) as \( 60 - x \). We want to maximize the product \( P = xy³ \). By substituting \( y \), the product becomes \( P(x) = x(60 - x)³ \). To find the maximum value, we first calculate the derivative of \( P(x) \) with respect to \( x \). We find \( \frac{dP}{dx} = (60 - x)²(60 - 4x) \). Setting this derivative to zero gives us \( x = 60 \) or \( x = 15 \). The value \( x = 60 \) is not possible as it would make \( y = 0 \), which is not a positive number. Using the second derivative test, or by observing the sign change of \( \frac{dP}{dx} \), we see that \( \frac{dP}{dx} \) changes from positive to negative as \( x \) increases through 15. This indicates a local maximum at \( x = 15 \). When \( x = 15 \), \( y = 60 - 15 = 45 \). Thus, the required positive numbers are 15 and 45.
In simple words: We need to find two numbers that add up to 60, and when the first number is multiplied by the cube of the second, the result is as large as possible. We found these numbers to be 15 and 45.

Exam Tip: When optimizing a product of variables with a fixed sum, express the product as a function of a single variable, then use differentiation to find critical points.

Question 15. Find two positive numbers x and y such that their sum is 35 and the product \( x²y⁵ \) is a maximum.
Answer: Let the two positive numbers be \( x \) and \( y \). We know their sum is 35, so \( x + y = 35 \), which means \( y = 35 - x \). We want to maximize the product \( P = x²y⁵ \). By substituting the expression for \( y \) into the product, we get \( P(x) = x²(35 - x)⁵ \). To find the maximum value, we compute the derivative of \( P(x) \) with respect to \( x \). Applying the product rule, we get \( \frac{dP}{dx} = x(35 - x)⁴(70 - 7x) \). Setting \( \frac{dP}{dx} = 0 \) yields \( x = 0 \), \( x = 35 \), or \( x = 10 \). Since \( x \) and \( y \) must be positive numbers, \( x = 0 \) and \( x = 35 \) are rejected because they would make either \( x \) or \( y \) zero. When \( x = 10 \), we check the sign change of \( \frac{dP}{dx} \). As \( x \) increases through 10, the derivative changes from positive to negative, indicating a maximum at \( x = 10 \). With \( x = 10 \), \( y = 35 - 10 = 25 \). So, the numbers are 10 and 25.
In simple words: We need two positive numbers that add up to 35. The goal is to make \( x \) squared times \( y \) to the power of five as big as it can be. The numbers that achieve this maximum product are 10 and 25.

Exam Tip: Always check boundary conditions and constraints (like "positive numbers") when finding maximum or minimum values to reject invalid solutions.

Question 16. Find two positive numbers whose sum is 16 and sum of whose cubes is minimum.
Answer: Let the two positive numbers be \( x \) and \( y \). Their sum is 16, so \( x + y = 16 \), which means \( y = 16 - x \). We want to minimize the sum of their cubes, \( S = x³ + y³ \). Substituting \( y \), we get \( S(x) = x³ + (16 - x)³ \). To find the minimum value, we compute the first derivative \( \frac{dS}{dx} \). We find \( \frac{dS}{dx} = 3x² - 3(16 - x)² \). Setting \( \frac{dS}{dx} = 0 \) gives \( x² = (16 - x)² \). Solving this, we get \( x = 16 - x \) or \( x = -(16 - x) \). The second case \( x = -16 + x \) gives \( 0 = -16 \), which is impossible. The first case gives \( 2x = 16 \), so \( x = 8 \). Now, we calculate the second derivative, \( \frac{d²S}{dx²} = 6x + 6(16 - x) = 6x + 96 - 6x = 96 \). Since \( \frac{d²S}{dx²} = 96 > 0 \), the value \( x = 8 \) corresponds to a minimum. If \( x = 8 \), then \( y = 16 - 8 = 8 \). Hence, the two numbers are 8 and 8.
In simple words: We are looking for two positive numbers that add up to 16. When you cube each number and then add those cubes together, the total should be the smallest possible. The two numbers that achieve this are 8 and 8.

Exam Tip: For problems involving sums and products, often the minimum or maximum occurs when the numbers are equal or close to equal, like in this case where \( x=y=8 \).

Question 17. 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 square to be cut off so that the volume of the box is maximum possible?
Answer: Let \( x \) be the side length of the square cut off from each corner. After cutting the squares and folding up the flaps, the dimensions of the open box will be: Length \( = 18 - 2x \) cm, Breadth \( = 18 - 2x \) cm, and Height \( = x \) cm. The volume of the box, \( V \), is given by \( V(x) = (18 - 2x)(18 - 2x)x = x(18 - 2x)² \). To find the maximum volume, we differentiate \( V(x) \) with respect to \( x \): \( \frac{dV}{dx} = (18 - 2x)² + x \cdot 2(18 - 2x)(-2) \). Simplifying this, we get \( \frac{dV}{dx} = (18 - 2x)(18 - 6x) \). Setting \( \frac{dV}{dx} = 0 \) gives \( 18 - 2x = 0 \) or \( 18 - 6x = 0 \). This yields \( x = 9 \) or \( x = 3 \). However, if \( x = 9 \), then the length and breadth would be \( 18 - 2(9) = 0 \), which is not possible for a box. So, \( x = 3 \) cm is the only viable option. To confirm it's a maximum, we calculate the second derivative \( \frac{d²V}{dx²} = (18 - 2x)(-6) + (18 - 6x)(-2) \). At \( x = 3 \), \( \frac{d²V}{dx²} = (18 - 6)(-6) + (18 - 18)(-2) = 12(-6) + 0 = -72 \). Since \( \frac{d²V}{dx²} < 0 \), the volume is maximum when \( x = 3 \) cm.
In simple words: To make the largest possible open box from a square piece of tin, you need to cut out squares from each corner. The side length of each small square you cut should be 3 cm.

Exam Tip: In optimization problems, ensure your answer makes physical sense; discard solutions that lead to impossible dimensions (e.g., zero length).

Question 18. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?
Answer: Let \( x \) be the side length of the square cut off from each corner. After cutting the squares and folding the flaps, the dimensions of the open box will be: Length \( = 45 - 2x \) cm, Breadth \( = 24 - 2x \) cm, and Height \( = x \) cm. The volume of the box, \( V \), is given by \( V(x) = x(45 - 2x)(24 - 2x) \). Expanding this, we get \( V(x) = 2x(2x² - 69x + 540) = 4x³ - 138x² + 1080x \). To find the maximum volume, we compute the first derivative \( \frac{dV}{dx} = 12x² - 276x + 1080 \). Setting \( \frac{dV}{dx} = 0 \) gives \( 12(x² - 23x + 90) = 0 \). Factoring the quadratic, we get \( 12(x - 5)(x - 18) = 0 \). This yields \( x = 5 \) or \( x = 18 \). However, if \( x = 18 \), the breadth would be \( 24 - 2(18) = 24 - 36 = -12 \), which is impossible. So, \( x = 5 \) cm is the only feasible option. To confirm it's a maximum, we calculate the second derivative \( \frac{d²V}{dx²} = 24x - 276 \). At \( x = 5 \), \( \frac{d²V}{dx²} = 24(5) - 276 = 120 - 276 = -156 \). Since \( \frac{d²V}{dx²} < 0 \), the volume is maximum when \( x = 5 \) cm.
In simple words: To make the largest open box from a rectangular piece of tin, you need to cut squares from each corner. The side length of the square to cut should be 5 cm.

Exam Tip: Always define your variables clearly and formulate the quantity to be maximized or minimized as a function of a single variable, respecting domain constraints.

Question 19. Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius a is a square side \( \sqrt{2}a \).
Answer: Let \( x \) and \( y \) be the length and breadth of the rectangle inscribed in a circle of radius \( a \). The diagonal of the rectangle will be the diameter of the circle, which is \( 2a \). So, by Pythagoras theorem, \( x² + y² = (2a)² = 4a² \). The perimeter of the rectangle is \( P = 2(x + y) \). We can express \( y \) in terms of \( x \) from the first equation: \( y = \sqrt{4a² - x²} \). Thus, \( P(x) = 2(x + \sqrt{4a² - x²}) \). To find the maximum perimeter, we differentiate \( P(x) \) with respect to \( x \): \( \frac{dP}{dx} = 2(1 - \frac{x}{\sqrt{4a² - x²}}) \). Setting \( \frac{dP}{dx} = 0 \) gives \( 1 = \frac{x}{\sqrt{4a² - x²}} \), which implies \( \sqrt{4a² - x²} = x \). Squaring both sides, \( 4a² - x² = x² \), so \( 4a² = 2x² \), leading to \( x² = 2a² \), or \( x = \sqrt{2}a \) (since length must be positive). To confirm it's a maximum, we examine the second derivative \( \frac{d²P}{dx²} \), which turns out to be negative at this value of \( x \). When \( x = \sqrt{2}a \), substituting back into \( y = \sqrt{4a² - x²} \) gives \( y = \sqrt{4a² - (\sqrt{2}a)²} = \sqrt{4a² - 2a²} = \sqrt{2a²} = \sqrt{2}a \). Since \( x = y \), the rectangle is a square with side length \( \sqrt{2}a \).
In simple words: For a rectangle drawn inside a circle, to make its boundary as long as possible (maximum perimeter), that rectangle must actually be a square. The side length of this square will be \( \sqrt{2} \) times the circle's radius.

Exam Tip: When dealing with inscribed figures in optimization problems, always relate the dimensions of the inscribed figure to the radius or diameter of the circle, usually through the Pythagorean theorem.

Question 20. Show that for a given surface area, the volume of a cylinder is maximum when its height is equal to the diameter of its base.
Answer: Let \( r \) be the radius and \( h \) be the height of the circular cylinder. Let \( S \) be its given surface area and \( V \) be its volume. The surface area is \( S = 2\pi r² + 2\pi rh \). From this, we can express \( h \) as \( h = \frac{S - 2\pi r²}{2\pi r} \). The volume of the cylinder is \( V = \pi r²h \). Substituting the expression for \( h \), we get \( V(r) = \pi r² \left( \frac{S - 2\pi r²}{2\pi r} \right) = \frac{1}{2}r(S - 2\pi r²) = \frac{1}{2}(Sr - 2\pi r³) \). To maximize the volume, we differentiate \( V(r) \) with respect to \( r \): \( \frac{dV}{dr} = \frac{1}{2}(S - 6\pi r²) \). Setting \( \frac{dV}{dr} = 0 \) gives \( S - 6\pi r² = 0 \), so \( S = 6\pi r² \). Now, we substitute this value of \( S \) back into the equation for \( h \): \( h = \frac{6\pi r² - 2\pi r²}{2\pi r} = \frac{4\pi r²}{2\pi r} = 2r \). To confirm this is a maximum, we compute the second derivative \( \frac{d²V}{dr²} = \frac{1}{2}(-12\pi r) = -6\pi r \). Since \( r \) is a radius, \( r > 0 \), which means \( -6\pi r < 0 \). Thus, \( V \) is maximum when \( h = 2r \), which means the height of the cylinder is equal to the diameter of its base.
In simple words: For a cylinder with a fixed amount of material for its surface (surface area), you get the largest volume when its height is the same as the width of its base (diameter).

Exam Tip: When a quantity is given as constant (like surface area here), use it to eliminate one variable in the expression you are optimizing, reducing it to a single-variable calculus problem.

Question 21. Of all the closed cylindrical cans (right circular cylinders) of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area.
Answer: Let \( r \) be the radius and \( h \) be the height of the cylindrical can. The given volume is \( V = \pi r²h = 100 \) cm³. From this, we can express \( h \) as \( h = \frac{100}{\pi r²} \). The total surface area of the closed can is \( S = 2\pi r² + 2\pi rh \). Substituting the expression for \( h \), we get \( S(r) = 2\pi r² + 2\pi r \left( \frac{100}{\pi r²} \right) = 2\pi r² + \frac{200}{r} \). To find the minimum surface area, we differentiate \( S(r) \) with respect to \( r \): \( \frac{dS}{dr} = 4\pi r - \frac{200}{r²} \). Setting \( \frac{dS}{dr} = 0 \) gives \( 4\pi r = \frac{200}{r²} \), so \( 4\pi r³ = 200 \), which implies \( \pi r³ = 50 \), or \( r = \left( \frac{50}{\pi} \right)^{\frac{1}{3}} \) cm. To confirm this is a minimum, we calculate the second derivative \( \frac{d²S}{dr²} = 4\pi + \frac{400}{r³} \). At \( r = \left( \frac{50}{\pi} \right)^{\frac{1}{3}} \), both terms are positive, so \( \frac{d²S}{dr²} > 0 \). This confirms that the surface area is minimum. Now, we find the height \( h \): \( h = \frac{100}{\pi r²} = \frac{100}{\pi \left( \frac{50}{\pi} \right)^{\frac{2}{3}}} = \frac{100 \cdot \pi^{\frac{2}{3}}}{\pi \cdot 50^{\frac{2}{3}}} = \frac{100}{50^{\frac{2}{3}} \cdot \pi^{\frac{1}{3}}} = 2 \left( \frac{50}{\pi} \right)^{\frac{1}{3}} \). So, \( h = 2r \). The dimensions for minimum surface area are \( r = \left( \frac{50}{\pi} \right)^{\frac{1}{3}} \) cm and \( h = 2 \left( \frac{50}{\pi} \right)^{\frac{1}{3}} \) cm.
In simple words: To make a cylindrical can with a certain amount of space (volume) but use the least amount of material (surface area), the height of the can should be twice its radius. This means the height should be the same as the diameter of its base.

Exam Tip: Remember that minimum surface area for a given volume in a cylinder also occurs when the height is equal to the diameter of the base \( (h = 2r) \).

Question 22. 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 lengths of the two pieces so that the combined area of the square and the circle is minimum?
Answer: Let the total length of the wire be 28 m. Suppose one piece has length \( x \) meters, which is used to form a circle. Then the circumference of the circle is \( x \), so \( 2\pi r = x \), which means its radius \( r = \frac{x}{2\pi} \). The area of the circle is \( A_{\text{circle}} = \pi r² = \pi \left( \frac{x}{2\pi} \right)² = \frac{x²}{4\pi} \). The other piece of wire has length \( (28 - x) \) meters, which is used to form a square. The perimeter of the square is \( (28 - x) \), so the side length of the square is \( s = \frac{28 - x}{4} \). The area of the square is \( A_{\text{square}} = s² = \left( \frac{28 - x}{4} \right)² \). The total combined area \( A(x) = A_{\text{circle}} + A_{\text{square}} = \frac{x²}{4\pi} + \left( \frac{28 - x}{4} \right)² \). To minimize the total area, we differentiate \( A(x) \) with respect to \( x \): \( \frac{dA}{dx} = \frac{2x}{4\pi} + 2 \left( \frac{28 - x}{4} \right) \left( -\frac{1}{4} \right) = \frac{x}{2\pi} - \frac{28 - x}{8} \). Setting \( \frac{dA}{dx} = 0 \) gives \( \frac{x}{2\pi} = \frac{28 - x}{8} \). Solving for \( x \), we get \( 8x = 2\pi(28 - x) \), which simplifies to \( 4x = 28\pi - \pi x \). Rearranging, \( (4 + \pi)x = 28\pi \), so \( x = \frac{28\pi}{4 + \pi} \) meters. The length of the second piece is \( 28 - x = 28 - \frac{28\pi}{4 + \pi} = \frac{28(4 + \pi) - 28\pi}{4 + \pi} = \frac{112}{4 + \pi} \) meters. To confirm this is a minimum, we find the second derivative \( \frac{d²A}{dx²} = \frac{1}{2\pi} + \frac{1}{8} \). Since this is always positive, the area is indeed minimized at this value of \( x \).
In simple words: We have a 28-meter wire to cut into two pieces. One piece forms a circle, and the other forms a square. We want the total area of both shapes to be as small as possible. The first piece (for the circle) should be \( \frac{28\pi}{4 + \pi} \) meters long, and the second piece (for the square) should be \( \frac{112}{4 + \pi} \) meters long.

Exam Tip: Be careful with algebra when setting up and solving derivatives for optimization problems, especially when involving constants like \( \pi \).

Question 23. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is \( \frac{8}{27} \) of the volume of the sphere.
Answer: Let \( R \) be the radius of the sphere. Consider a cone inscribed in the sphere. Let the semi-vertical angle of the cone be \( \theta \). Then the radius of the cone's base is \( r = R \sin\theta \) and its height is \( h = R + R \cos\theta \). The volume of the cone is given by \( V = \frac{1}{3}\pi r²h \). Substituting the expressions for \( r \) and \( h \), we get \( V(\theta) = \frac{1}{3}\pi (R \sin\theta)² (R + R \cos\theta) = \frac{1}{3}\pi R³ \sin²\theta (1 + \cos\theta) \). To find the maximum volume, we differentiate \( V(\theta) \) with respect to \( \theta \): \( \frac{dV}{d\theta} = \frac{1}{3}\pi R³ [2\sin\theta\cos\theta(1+\cos\theta) + \sin²\theta(-\sin\theta)] \). This simplifies to \( \frac{dV}{d\theta} = \frac{1}{3}\pi R³ \sin\theta (3\cos²\theta + 2\cos\theta - 1) \). Setting \( \frac{dV}{d\theta} = 0 \) (and since \( \sin\theta \neq 0 \) for a cone), we solve \( 3\cos²\theta + 2\cos\theta - 1 = 0 \). Factoring this quadratic in \( \cos\theta \), we get \( (3\cos\theta - 1)(\cos\theta + 1) = 0 \). This yields \( \cos\theta = \frac{1}{3} \) or \( \cos\theta = -1 \). The value \( \cos\theta = -1 \) implies \( \theta = \pi \), which would mean the cone is flat and has no volume, so we reject it. Thus, \( \cos\theta = \frac{1}{3} \). When \( \cos\theta = \frac{1}{3} \), we can find \( \sin²\theta = 1 - \cos²\theta = 1 - (\frac{1}{3})² = 1 - \frac{1}{9} = \frac{8}{9} \). Substituting these values back into the volume formula for the cone: \( V_{\text{max}} = \frac{1}{3}\pi R³ \left( \frac{8}{9} \right) \left( 1 + \frac{1}{3} \right) = \frac{1}{3}\pi R³ \left( \frac{8}{9} \right) \left( \frac{4}{3} \right) = \frac{32}{81}\pi R³ \). The volume of the sphere is \( V_{\text{sphere}} = \frac{4}{3}\pi R³ \). Comparing \( V_{\text{max}} \) with \( V_{\text{sphere}} \): \( \frac{V_{\text{max}}}{V_{\text{sphere}}} = \frac{\frac{32}{81}\pi R³}{\frac{4}{3}\pi R³} = \frac{32}{81} \times \frac{3}{4} = \frac{8 \times 1}{27 \times 1} = \frac{8}{27} \). Hence, the maximum volume of the inscribed cone is \( \frac{8}{27} \) of the volume of the sphere.
In simple words: When you fit the biggest possible cone inside a sphere, the cone's volume will always be \( \frac{8}{27} \) of the sphere's total volume. This happens when the cosine of the cone's half-angle is \( \frac{1}{3} \).

Exam Tip: For optimization problems with geometric figures, draw a clear diagram and relate the variables using geometry (Pythagoras, trigonometry) before applying calculus.

Question 24. Show that the right circular cone of least curved surface area and given volume has an altitude equal to \( \sqrt{2} \) times the radius of the base.
Answer: Let \( r \) be the radius of the base and \( h \) be the altitude (height) of the right circular cone. Let \( l \) be its slant height, so \( l = \sqrt{r² + h²} \). The volume \( V \) of the cone is given as a constant, \( V = \frac{1}{3}\pi r²h \). From this, we can write \( h = \frac{3V}{\pi r²} \). The curved surface area of the cone is \( S = \pi rl = \pi r\sqrt{r² + h²} \). To minimize \( S \), it's equivalent to minimize \( S² \). Let \( Z = S² = \pi² r²(r² + h²) \). Substitute \( h \): \( Z(r) = \pi² r² \left( r² + \left( \frac{3V}{\pi r²} \right)² \right) = \pi² r² \left( r² + \frac{9V²}{\pi² r⁴} \right) = \pi² r⁴ + \frac{9V²}{r²} \). To find the minimum \( Z \), we differentiate with respect to \( r \): \( \frac{dZ}{dr} = 4\pi² r³ - \frac{18V²}{r³} \). Setting \( \frac{dZ}{dr} = 0 \) gives \( 4\pi² r³ = \frac{18V²}{r³} \), so \( 4\pi² r⁶ = 18V² \), which simplifies to \( 2\pi² r⁶ = 9V² \). Now substitute \( V = \frac{1}{3}\pi r²h \): \( 2\pi² r⁶ = 9 \left( \frac{1}{3}\pi r²h \right)² = 9 \left( \frac{1}{9}\pi² r⁴h² \right) = \pi² r⁴h² \). Dividing by \( \pi² r⁴ \) (since \( r \neq 0 \)), we get \( 2r² = h² \), which means \( h = \sqrt{2}r \). To confirm this is a minimum, we find the second derivative \( \frac{d²Z}{dr²} = 12\pi² r² + \frac{54V²}{r⁴} \). Since \( r \) and \( V \) are positive, \( \frac{d²Z}{dr²} > 0 \), confirming that the curved surface area is minimum when \( h = \sqrt{2}r \).
In simple words: To make a cone that holds a certain amount (volume) but uses the smallest possible amount of material for its slanted surface, the cone's height needs to be \( \sqrt{2} \) times the radius of its base.

Exam Tip: Sometimes, minimizing the square of a quantity (like \( S² \) instead of \( S \)) simplifies differentiation significantly without changing the location of the extremum.

Question 25. Show that the semi-vertical angle of the cone of maximum volume and of given slant height is \( \tan^{-1}\sqrt{2} \).
Answer: Let \( l \) be the given slant height of the cone, which is constant. Let \( r \) be the radius of the base and \( h \) be the vertical height of the cone. Let \( \theta \) be the semi-vertical angle. From trigonometry, \( r = l \sin\theta \) and \( h = l \cos\theta \). The volume of the cone is \( V = \frac{1}{3}\pi r²h \). Substituting the expressions for \( r \) and \( h \) in terms of \( l \) and \( \theta \), we get \( V(\theta) = \frac{1}{3}\pi (l \sin\theta)² (l \cos\theta) = \frac{1}{3}\pi l³ \sin²\theta \cos\theta \). To find the maximum volume, we differentiate \( V(\theta) \) with respect to \( \theta \): \( \frac{dV}{d\theta} = \frac{1}{3}\pi l³ [2\sin\theta\cos\theta \cdot \cos\theta + \sin²\theta (-\sin\theta)] = \frac{1}{3}\pi l³ [2\sin\theta\cos²\theta - \sin³\theta] \). Factoring out \( \sin\theta \), we get \( \frac{dV}{d\theta} = \frac{1}{3}\pi l³ \sin\theta (2\cos²\theta - \sin²\theta) \). Setting \( \frac{dV}{d\theta} = 0 \) (and since \( \sin\theta \neq 0 \) for a cone), we solve \( 2\cos²\theta - \sin²\theta = 0 \). This implies \( 2\cos²\theta = \sin²\theta \), so \( 2 = \frac{\sin²\theta}{\cos²\theta} = \tan²\theta \). Taking the square root, \( \tan\theta = \sqrt{2} \) (since \( \theta \) is an acute angle). Therefore, the semi-vertical angle \( \theta = \tan^{-1}\sqrt{2} \). To confirm this is a maximum, we can use the second derivative test, which shows \( \frac{d²V}{d\theta²} < 0 \) at this value of \( \theta \), indicating a maximum volume.
In simple words: If you have a cone with a specific slanted edge length, to make its volume as large as possible, its semi-vertical angle (half the angle at the tip) must be such that its tangent is \( \sqrt{2} \).

Exam Tip: When given a constant slant height for a cone, it's often easiest to express its radius and height using trigonometric functions of the semi-vertical angle.

Question 26. Show that semi-vertical angle of a right circular cone of given surface area and maximum volume is \( \sin^{-1}\frac{1}{3} \).
Answer: Let \( r \) be the radius of the base, \( h \) be the height, and \( l \) be the slant height of the right circular cone. Let \( \alpha \) be the semi-vertical angle. The total surface area \( S \) is given as a constant, \( S = \pi r² + \pi rl \). From this, we can express the slant height as \( l = \frac{S - \pi r²}{\pi r} \). The volume of the cone is \( V = \frac{1}{3}\pi r²h \). We also know that \( h = \sqrt{l² - r²} \). Substituting \( h \), the volume becomes \( V = \frac{1}{3}\pi r²\sqrt{l² - r²} \). To maximize \( V \), we can maximize \( V² \). Let \( Z = V² = \frac{1}{9}\pi² r⁴(l² - r²) \). Now, substitute the expression for \( l \) into \( Z \): \( Z(r) = \frac{1}{9}\pi² r⁴ \left[ \left( \frac{S - \pi r²}{\pi r} \right)² - r² \right] = \frac{1}{9} r² ( (S - \pi r²)² - \pi² r⁴ ) = \frac{1}{9} r² (S² - 2\pi Sr² + \pi² r⁴ - \pi² r⁴) = \frac{S}{9}(Sr² - 2\pi r⁴) \). To find the maximum \( Z \), we differentiate with respect to \( r \): \( \frac{dZ}{dr} = \frac{S}{9}(2Sr - 8\pi r³) \). Setting \( \frac{dZ}{dr} = 0 \) gives \( 2Sr - 8\pi r³ = 0 \). Since \( r \neq 0 \), we have \( 2S = 8\pi r² \), so \( S = 4\pi r² \). To confirm this is a maximum, the second derivative \( \frac{d²Z}{dr²} \) will be negative at this value. Now we relate this back to the semi-vertical angle \( \alpha \). We have \( S = 4\pi r² \) and \( S = \pi r² + \pi rl \). Equating these, \( 4\pi r² = \pi r² + \pi rl \). Dividing by \( \pi r \) (since \( r \neq 0 \)), we get \( 4r = r + l \), which implies \( l = 3r \). For a cone, the semi-vertical angle \( \alpha \) satisfies \( \sin\alpha = \frac{r}{l} \). Substituting \( l = 3r \), we get \( \sin\alpha = \frac{r}{3r} = \frac{1}{3} \). Therefore, the semi-vertical angle is \( \alpha = \sin^{-1}\frac{1}{3} \).
In simple words: For a cone with a given total surface area, to achieve the largest possible volume, its semi-vertical angle should be such that its sine value is \( \frac{1}{3} \). This means the slant height is three times the base radius.

Exam Tip: When surface area is given, it often involves both \( r \) and \( l \). Use the relation \( l² = r² + h² \) or \( \sin\alpha = r/l \) to link these dimensions to the optimization process.

Question 27. The point on the curve \( x² = 2y \) which is nearest to the point \( (0, 5) \) is
(A) \( (2\sqrt{2}, 4) \)
(B) \( (2\sqrt{2}, 0) \)
(C) (0, 0)
Answer: (A) \( (2\sqrt{2}, 4) \)
Let \( P(x, y) \) be a point on the curve \( x² = 2y \). The given point is \( A(0, 5) \). To find the nearest point, we need to minimize the distance between \( P \) and \( A \). Minimizing the distance is equivalent to minimizing the square of the distance, \( Z = PA² \). So, \( Z = (x - 0)² + (y - 5)² = x² + (y - 5)² \). Since \( x² = 2y \), we substitute this into the equation for \( Z \): \( Z(y) = 2y + (y - 5)² = 2y + y² - 10y + 25 = y² - 8y + 25 \). To minimize \( Z \), we differentiate \( Z(y) \) with respect to \( y \): \( \frac{dZ}{dy} = 2y - 8 \). Setting \( \frac{dZ}{dy} = 0 \) gives \( 2y - 8 = 0 \), so \( y = 4 \). To confirm this is a minimum, we find the second derivative \( \frac{d²Z}{dy²} = 2 \). Since \( \frac{d²Z}{dy²} > 0 \), \( Z \) is minimum at \( y = 4 \). Now, substitute \( y = 4 \) back into the curve's equation \( x² = 2y \): \( x² = 2(4) = 8 \). This means \( x = \pm \sqrt{8} = \pm 2\sqrt{2} \). So the points are \( (2\sqrt{2}, 4) \) and \( (-2\sqrt{2}, 4) \). Among the given options, \( (2\sqrt{2}, 4) \) is present.
In simple words: To find the spot on the curve \( x² = 2y \) that is closest to the point \( (0, 5) \), we used math to find the minimum distance. The closest point is \( (2\sqrt{2}, 4) \).

Exam Tip: When minimizing distance, it's often simpler to minimize the square of the distance to avoid square roots in the differentiation process.

Question 28. For all real values of x, the minimum value of \( \frac{1-x+x^{2}}{1+x+x^{2}} \) is
(A) 0
(B) 1
(C) 3
(D) \( \frac{1}{3} \)
Answer: (D) \( \frac{1}{3} \)
Let \( y = \frac{x² - x + 1}{x² + x + 1} \). To find the minimum value of \( y \), we can use two methods. Method 1 (Discriminant): Rearrange the expression to form a quadratic equation in \( x \). \( y(x² + x + 1) = x² - x + 1 \) \( \implies yx² + yx + y = x² - x + 1 \) \( \implies (y - 1)x² + (y + 1)x + (y - 1) = 0 \). For \( x \) to be a real number, the discriminant \( D \) of this quadratic must be greater than or equal to zero. So, \( D = (y + 1)² - 4(y - 1)(y - 1) \geq 0 \). \( (y + 1)² - 4(y - 1)² \geq 0 \). Expanding and simplifying, or using the difference of squares, we get \( (y+1 - 2(y-1))(y+1 + 2(y-1)) \geq 0 \). This simplifies to \( (-y + 3)(3y - 1) \geq 0 \). Multiplying by \( -1 \) and reversing the inequality, we get \( (y - 3)(3y - 1) \leq 0 \). This inequality holds when \( \frac{1}{3} \leq y \leq 3 \). Thus, the minimum value of \( y \) is \( \frac{1}{3} \). Method 2 (Calculus): Find \( \frac{dy}{dx} \). Using the quotient rule, \( \frac{dy}{dx} = \frac{(2x - 1)(x² + x + 1) - (x² - x + 1)(2x + 1)}{(x² + x + 1)²} \). The numerator simplifies to \( 2x² - 2 \). So, \( \frac{dy}{dx} = \frac{2x² - 2}{(x² + x + 1)²} \). Setting \( \frac{dy}{dx} = 0 \) gives \( 2x² - 2 = 0 \implies x² = 1 \implies x = \pm 1 \). When \( x = 1 \), \( y = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3} \). When \( x = -1 \), \( y = \frac{1 - (-1) + 1}{1 + (-1) + 1} = \frac{3}{1} = 3 \). Using the second derivative test, \( x = 1 \) corresponds to a minimum, and \( x = -1 \) corresponds to a maximum. Therefore, the minimum value of the expression is \( \frac{1}{3} \).
In simple words: We want to find the smallest number this fraction can be. By using mathematical techniques, we found that the lowest value the fraction \( \frac{1-x+x^{2}}{1+x+x^{2}} \) can reach is \( \frac{1}{3} \).

Exam Tip: For rational functions of quadratics, both the calculus method (finding critical points via derivatives) and algebraic method (using the discriminant for real roots) are effective for determining the range of values.

Question 28. For all real values of x, the minimum value of \( \frac{1-x+x^{2}}{1+x+x^{2}} \) is
(a) 0
(b) 1
(c) 3
(d) \( \frac { 1 }{ 3 } \)
Answer: (d) \( \frac { 1 }{ 3 } \)
Let the function be \( y = \frac{1-x+x^{2}}{1+x+x^{2}} \). To find the minimum value, we need to calculate the first derivative \( \frac{dy}{dx} \).
The derivative is found as: \( \frac{dy}{dx} = \frac{(-1+2x)(1+ x + x^{2})-(1-x+x^{2})(1+2x)}{(1+x+x^{2})^{2}} \).
Simplifying the numerator, we get: \( (-1+2x)(1 + x + x^{2}) - (1 + 2x)(1-x+x^{2}) \).
This further simplifies to \( (-1 - x - x^{2}) + (2x + 2x^{2}+2x^{3}) - (1-x + x^{2})-(2x-2x^{2} + 2x^{3}) \).
After combining terms, the numerator becomes \( -2 + 2x^{2} \), which can be written as \( 2(x^{2} - 1) \) or \( 2(x - 1)(x + 1) \).
So, \( \frac{dy}{dx} = \frac{2(x-1)(x+1)}{(x^{2}+x+1)^{2}} \).
Setting \( \frac{dy}{dx} = 0 \) gives us \( x = 1 \) or \( x = -1 \).
When \( x = 1 \), the sign of \( \frac{dy}{dx} \) changes from negative to positive. This shows that the function has a local minimum at \( x = 1 \).
The minimum value of the function at \( x = 1 \) is found by substituting \( x=1 \) into the original function: \( y = \frac{1-1+1}{1+1+1} = \frac{1}{3} \).
Therefore, the correct choice is (d).
In simple words: To find the smallest value, we first take the derivative of the given function and set it to zero. This helps us find the critical points. When we test the point \( x=1 \), we see that the function changes from decreasing to increasing, which means it's a minimum. Plugging \( x=1 \) back into the original function gives us \( \frac{1}{3} \).

Exam Tip: For rational functions, always simplify the derivative carefully and check the sign change around critical points to confirm if it's a maximum or minimum.

Question 29. The maximum value of \( (x(x-1)+1)^{\frac{1}{3}} \), for \( 0 \leq x \leq 1 \) is
(a) \( (\frac { 1 }{ 3 })^{\frac { 1 }{ 3 }} \)
(b) \( \frac { 1 }{ 2 } \)
(c) 1
(d) \( \frac { 1 }{ 3 } \)
Answer: (c) 1
Let the given function be \( y = (x(x-1)+1)^{\frac{1}{3}} \), which can be written as \( y = (x^2 - x + 1)^{\frac{1}{3}} \). We need to find its maximum value in the interval \( [0, 1] \).
First, find the derivative \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{1}{3} (x^2 - x + 1)^{-\frac{2}{3}} (2x - 1) \).
To find critical points, set \( \frac{dy}{dx} = 0 \). This gives \( 2x - 1 = 0 \), so \( x = \frac{1}{2} \).
The sign of \( \frac{dy}{dx} \) changes from negative to positive as \( x \) passes through \( \frac{1}{2} \), indicating a local minimum at this point.
Now, we evaluate the function at the critical point and at the endpoints of the interval \( [0, 1] \):
At \( x = \frac{1}{2} \): \( y = ((\frac{1}{2})^2 - \frac{1}{2} + 1)^{\frac{1}{3}} = (\frac{1}{4} - \frac{1}{2} + 1)^{\frac{1}{3}} = (\frac{3}{4})^{\frac{1}{3}} \).
At \( x = 0 \): \( y = (0^2 - 0 + 1)^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1 \).
At \( x = 1 \): \( y = (1^2 - 1 + 1)^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1 \).
Comparing these values, \( (\frac{3}{4})^{\frac{1}{3}} \approx 0.908 \) which is less than 1. Therefore, the maximum value of \( y \) in the given interval is 1.
The correct option is (c).
In simple words: We want to find the biggest value of the function. First, we calculate its derivative and set it to zero to find the points where it might be a maximum or minimum. We find a minimum at \( x = \frac{1}{2} \). Then, we check the function's value at this point and at the ends of the given range (from 0 to 1). The biggest value we find is 1.

Exam Tip: To find absolute maximum or minimum values over a closed interval, always check the function's values at both critical points within the interval and at the endpoints of the interval.