Rbse Solutions Class 12 Maths Chapter 6 Exercise 6.3 | Maxima and Minima

Rbse Solutions Class 12 Maths Chapter 6 Exercise 6.3 | Maxima and Minima

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Complete solutions for Class 12 Maths (NCERT) Exercise 6.3. Solve problems on Local Maxima and Minima (First and Second Derivative Tests), Absolute Extrema on closed intervals, and complex Optimization problems for geometry (e.g., box volume, cylinder surface area, cone volume).

image 123 Rbse Solutions Class 12 Maths Chapter 6 Exercise 6.3 | Maxima and Minima

1. Global Maxima and Minima (Basic Functions)

(i) $f(x) = (2x – 1)^2 + 3$

Since $(2x – 1)^2 \ge 0$ for all $x \in \mathbb{R}$,

  • The minimum value of $(2x – 1)^2$ is 0, which occurs when $2x – 1 = 0 \implies x = 1/2$.
  • Minimum Value: $f(1/2) = 0 + 3 = \mathbf{3}$.
  • Maximum Value: As $x \to \pm \infty$, $f(x) \to \infty$. Does not exist.

(ii) $f(x) = 9x^2 + 12x + 2$

Complete the square:

$$f(x) = (9x^2 + 12x + 4) – 4 + 2 = (3x + 2)^2 – 2$$

Since $(3x + 2)^2 \ge 0$ for all $x \in \mathbb{R}$,

  • The minimum value is 0, when $3x + 2 = 0 \implies x = -2/3$.
  • Minimum Value: $f(-2/3) = 0 – 2 = \mathbf{-2}$.
  • Maximum Value: As $x \to \pm \infty$, $f(x) \to \infty$. Does not exist.

(iii) $f(x) = – (x – 1)^2 + 10$

Since $-(x – 1)^2 \le 0$ for all $x \in \mathbb{R}$,

  • The maximum value of $-(x – 1)^2$ is 0, which occurs when $x – 1 = 0 \implies x = 1$.
  • Maximum Value: $f(1) = 0 + 10 = \mathbf{10}$.
  • Minimum Value: As $x \to \pm \infty$, $f(x) \to -\infty$. Does not exist.

(iv) $g(x) = x^3 + 1$

  • As $x \to \infty$, $g(x) \to \infty$. Maximum value does not exist.
  • As $x \to -\infty$, $g(x) \to -\infty$. Minimum value does not exist.

2. Global Maxima and Minima (Modulus and Trigonometric Functions)

(i) $f(x) = |x + 2| – 1$

Since $|x + 2| \ge 0$ for all $x \in \mathbb{R}$,

  • The minimum value is 0, when $x + 2 = 0 \implies x = -2$.
  • Minimum Value: $f(-2) = 0 – 1 = \mathbf{-1}$.
  • Maximum Value: As $x \to \pm \infty$, $f(x) \to \infty$. Does not exist.

(ii) $g(x) = – |x + 1| + 3$

Since $-|x + 1| \le 0$ for all $x \in \mathbb{R}$,

  • The maximum value is 0, when $x + 1 = 0 \implies x = -1$.
  • Maximum Value: $g(-1) = 0 + 3 = \mathbf{3}$.
  • Minimum Value: As $x \to \pm \infty$, $g(x) \to -\infty$. Does not exist.

(iii) $h(x) = \sin(2x) + 5$

We know that $-1 \le \sin(2x) \le 1$ for all $x \in \mathbb{R}$.

  • Maximum Value: $1 + 5 = \mathbf{6}$.
  • Minimum Value: $-1 + 5 = \mathbf{4}$.

(iv) $f(x) = |\sin 4x + 3|$

We know that $-1 \le \sin 4x \le 1$.

  • $-1 + 3 \le \sin 4x + 3 \le 1 + 3$
  • $2 \le \sin 4x + 3 \le 4$Since $\sin 4x + 3$ is always positive, $| \sin 4x + 3 | = \sin 4x + 3$.
  • Maximum Value: $|\mathbf{4}| = \mathbf{4}$.
  • Minimum Value: $|\mathbf{2}| = \mathbf{2}$.

(v) $h(x) = x + 1, x \in (-1, 1)$

The function is continuous on the open interval $(-1, 1)$.

  • As $x \to 1^-$, $h(x) \to 1 + 1 = 2$. Maximum value does not exist.
  • As $x \to -1^+$, $h(x) \to -1 + 1 = 0$. Minimum value does not exist.

3. Local Maxima and Minima

(i) $f(x) = x^2$

  1. $f'(x) = 2x$. Critical point: $f'(x) = 0 \implies x = 0$.
  2. $f”(x) = 2$.
  3. $f”(0) = 2 > 0$. $\implies$ Local Minima at $x=0$.
  4. Local Minimum Value: $f(0) = \mathbf{0}$.

(ii) $g(x) = x^3 – 3x$

  1. $g'(x) = 3x^2 – 3 = 3(x^2 – 1)$. Critical points: $g'(x) = 0 \implies x = \pm 1$.
  2. $g”(x) = 6x$.
  3. $g”(1) = 6(1) = 6 > 0$. $\implies$ Local Minima at $x=1$.
  4. $g”(-1) = 6(-1) = -6 < 0$. $\implies$ Local Maxima at $x=-1$.
  5. Local Maximum Value: $g(-1) = (-1)^3 – 3(-1) = -1 + 3 = \mathbf{2}$.
  6. Local Minimum Value: $g(1) = (1)^3 – 3(1) = 1 – 3 = \mathbf{-2}$.

(iii) $h(x) = \sin x + \cos x, 0 < x < \frac{\pi}{2}$

  1. $h'(x) = \cos x – \sin x$. Critical point: $\cos x = \sin x \implies \tan x = 1$. Since $x \in (0, \pi/2)$, $x = \pi/4$.
  2. $h”(x) = -\sin x – \cos x$.
  3. $h”(\pi/4) = -\sin(\pi/4) – \cos(\pi/4) = -\frac{1}{\sqrt{2}} – \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} < 0$. $\implies$ Local Maxima at $x=\pi/4$.
  4. Local Maximum Value: $h(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \mathbf{\sqrt{2}}$.

(iv) $f(x) = \sin x – \cos x, 0 < x < 2\pi$

  1. $f'(x) = \cos x + \sin x$. Critical points: $\cos x = -\sin x \implies \tan x = -1$. In $[0, 2\pi]$, $x = 3\pi/4$ and $x = 7\pi/4$.
  2. $f”(x) = -\sin x + \cos x$.
  3. $f”(3\pi/4) = -\sin(3\pi/4) + \cos(3\pi/4) = -\frac{1}{\sqrt{2}} – \frac{1}{\sqrt{2}} < 0$. $\implies$ Local Maxima at $x=3\pi/4$.
  4. $f”(7\pi/4) = -\sin(7\pi/4) + \cos(7\pi/4) = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} > 0$. $\implies$ Local Minima at $x=7\pi/4$.
  5. Local Maximum Value: $f(3\pi/4) = \sin(3\pi/4) – \cos(3\pi/4) = \frac{1}{\sqrt{2}} – (-\frac{1}{\sqrt{2}}) = \mathbf{\sqrt{2}}$.
  6. Local Minimum Value: $f(7\pi/4) = \sin(7\pi/4) – \cos(7\pi/4) = -\frac{1}{\sqrt{2}} – \frac{1}{\sqrt{2}} = \mathbf{-\sqrt{2}}$.

(v) $f(x) = x^3 – 6x^2 + 9x + 15$

  1. $f'(x) = 3x^2 – 12x + 9 = 3(x^2 – 4x + 3) = 3(x – 1)(x – 3)$. Critical points: $x = 1, 3$.
  2. $f”(x) = 6x – 12$.
  3. $f”(1) = 6(1) – 12 = -6 < 0$. $\implies$ Local Maxima at $x=1$.
  4. $f”(3) = 6(3) – 12 = 6 > 0$. $\implies$ Local Minima at $x=3$.
  5. Local Maximum Value: $f(1) = 1 – 6 + 9 + 15 = \mathbf{19}$.
  6. Local Minimum Value: $f(3) = 27 – 6(9) + 9(3) + 15 = 27 – 54 + 27 + 15 = \mathbf{15}$.

(vi) $g(x) = \frac{x}{2} + \frac{2}{x}, x > 0$

  1. $g'(x) = \frac{1}{2} – \frac{2}{x^2}$. Critical points: $\frac{1}{2} = \frac{2}{x^2} \implies x^2 = 4$. Since $x > 0$, $x = 2$.
  2. $g”(x) = \frac{4}{x^3}$.
  3. $g”(2) = \frac{4}{2^3} = \frac{1}{2} > 0$. $\implies$ Local Minima at $x=2$.
  4. Local Minimum Value: $g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = \mathbf{2}$.

(vii) $g(x) = \frac{1}{x^2 + 2}$

  1. $g'(x) = -\frac{2x}{(x^2 + 2)^2}$. Critical point: $g'(x) = 0 \implies x = 0$.
  2. $g'(x)$ changes from positive (when $x < 0$) to negative (when $x > 0$). $\implies$ Local Maxima at $x=0$.
  3. Local Maximum Value: $g(0) = \frac{1}{0 + 2} = \mathbf{1/2}$.

(viii) $f(x) = x \sqrt{1-x}, 0 < x < 1$

  1. $f'(x) = 1 \cdot \sqrt{1-x} + x \cdot \frac{1}{2\sqrt{1-x}} (-1) = \sqrt{1-x} – \frac{x}{2\sqrt{1-x}}$$$f'(x) = \frac{2(1-x) – x}{2\sqrt{1-x}} = \frac{2 – 3x}{2\sqrt{1-x}}$$
  2. Critical point: $2 – 3x = 0 \implies x = 2/3$.
  3. Check sign change around $x=2/3$:
    • For $x < 2/3$ (e.g., $x=0.5$): $2 – 3x > 0$. $f'(x) > 0$. (Increasing)
    • For $x > 2/3$ (e.g., $x=0.8$): $2 – 3x < 0$. $f'(x) < 0$. (Decreasing)$\implies$ Local Maxima at $x=2/3$.
  4. Local Maximum Value: $f(2/3) = \frac{2}{3} \sqrt{1 – \frac{2}{3}} = \frac{2}{3} \sqrt{\frac{1}{3}} = \frac{2}{3\sqrt{3}} = \mathbf{\frac{2\sqrt{3}}{9}}$.

4. Functions Without Maxima or Minima

For a continuous function to have local maxima/minima, its derivative must be zero at some point.

(i) $f(x) = e^x$

$f'(x) = e^x$. Since $e^x > 0$ for all $x \in \mathbb{R}$, $f'(x)$ is never zero. The function is strictly increasing throughout $\mathbb{R}$. It has no local maxima or minima.

(ii) $g(x) = \log x$

$g'(x) = \frac{1}{x}$. Since $x > 0$ (domain of $\log x$), $g'(x) > 0$. $g'(x)$ is never zero. The function is strictly increasing throughout its domain $(0, \infty)$. It has no local maxima or minima.

(iii) $h(x) = x^3 + x^2 + x + 1$

$h'(x) = 3x^2 + 2x + 1$.

The discriminant of this quadratic is $D = b^2 – 4ac = 2^2 – 4(3)(1) = 4 – 12 = -8$.

Since $D < 0$ and the leading coefficient ($3$) is positive, $h'(x) > 0$ for all $x \in \mathbb{R}$. $h'(x)$ is never zero. The function is strictly increasing throughout $\mathbb{R}$. It has no local maxima or minima.

5. Absolute Maxima and Minima (Closed Intervals)

The absolute extrema occur either at the critical points within the interval or at the endpoints.

(i) $f(x) = x^3, x \in [-2, 2]$

  1. $f'(x) = 3x^2$. Critical point: $x = 0$.
  2. Evaluate:
    • $f(-2) = (-2)^3 = -8$
    • $f(0) = 0$
    • $f(2) = (2)^3 = 8$
  3. Absolute Maximum Value: $\mathbf{8}$ at $x=2$.
  4. Absolute Minimum Value: $\mathbf{-8}$ at $x=-2$.

(ii) $f(x) = \sin x + \cos x, x \in [0, \pi]$

  1. $f'(x) = \cos x – \sin x$. Critical point: $\cos x = \sin x \implies \tan x = 1$. In $[0, \pi]$, $x = \pi/4$.
  2. Evaluate:
    • $f(0) = \sin 0 + \cos 0 = 0 + 1 = 1$
    • $f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414$
    • $f(\pi) = \sin \pi + \cos \pi = 0 – 1 = -1$
  3. Absolute Maximum Value: $\mathbf{\sqrt{2}}$ at $x=\pi/4$.
  4. Absolute Minimum Value: $\mathbf{-1}$ at $x=\pi$.

(iii) $f(x) = 12x^{4/3} – 6x^{1/3}, x \in [-1, 1]$

  1. $f'(x) = 12 \cdot \frac{4}{3} x^{1/3} – 6 \cdot \frac{1}{3} x^{-2/3} = 16x^{1/3} – 2x^{-2/3}$$$f'(x) = 2x^{-2/3} (8x – 1) = \frac{2(8x – 1)}{x^{2/3}}$$
  2. Critical points:
    • $f'(x) = 0 \implies 8x – 1 = 0 \implies x = 1/8$.
    • $f'(x)$ is undefined at $x = 0$.
  3. Evaluate:
    • $f(-1) = 12(-1)^{4/3} – 6(-1)^{1/3} = 12(1) – 6(-1) = 18$
    • $f(1/8) = 12(\frac{1}{8})^{4/3} – 6(\frac{1}{8})^{1/3} = 12(\frac{1}{16}) – 6(\frac{1}{2}) = \frac{3}{4} – 3 = -\frac{9}{4} = -2.25$
    • $f(0) = 0 – 0 = 0$
    • $f(1) = 12(1) – 6(1) = 6$
  4. Absolute Maximum Value: $\mathbf{18}$ at $x=-1$.
  5. Absolute Minimum Value: $\mathbf{-9/4}$ at $x=1/8$.

(iv) $f(x) = (x – 1)^2 + 3, x \in [-3, 1]$

(Note: Question says $f(x) = x^2(x – 1)^2 + 3$, which is incorrect based on the typical problem structure. Assuming the function is $f(x) = (x – 1)^2 + 3$, as per the bracket structure of the original equation.)

  1. $f'(x) = 2(x – 1)$. Critical point: $x = 1$. (Endpoint)
  2. Evaluate:
    • $f(-3) = (-3 – 1)^2 + 3 = 16 + 3 = 19$
    • $f(1) = (1 – 1)^2 + 3 = 3$
  3. Absolute Maximum Value: $\mathbf{19}$ at $x=-3$.
  4. Absolute Minimum Value: $\mathbf{3}$ at $x=1$.

6-12. Optimization Problems (Calculus)

6. Maximum Profit

Profit function: $p(x) = 41 – 72x – 18x^2$.

  1. $p'(x) = -72 – 36x$. Critical point: $-72 – 36x = 0 \implies 36x = -72 \implies x = -2$.
  2. $p”(x) = -36$. Since $p”(-2) < 0$, it is a maximum.
  3. Maximum Profit: $p(-2) = 41 – 72(-2) – 18(-2)^2 = 41 + 144 – 18(4) = 185 – 72 = \mathbf{113}$.

7. Max/Min of $f(x) = 3x^4 – 8x^3 + 12x^2 – 48x + 25$ on $[0, 3]$

  1. $f'(x) = 12x^3 – 24x^2 + 24x – 48 = 12[x^2(x – 2) + 2(x – 2)] = 12(x – 2)(x^2 + 2)$.
  2. Critical point: $x – 2 = 0 \implies x = 2$. ($x^2 + 2$ is never zero).
  3. Evaluate:
    • $f(0) = 25$
    • $f(2) = 3(16) – 8(8) + 12(4) – 48(2) + 25 = 48 – 64 + 48 – 96 + 25 = -39$
    • $f(3) = 3(81) – 8(27) + 12(9) – 48(3) + 25 = 243 – 216 + 108 – 144 + 25 = 16$
  4. Maximum Value: $\mathbf{25}$ at $x=0$.
  5. Minimum Value: $\mathbf{-39}$ at $x=2$.

8. Maximum value of $\sin 2x$ on $[0, 2\pi]$

The maximum value of $\sin \theta$ is 1. We require $\sin 2x = 1$.

$$2x = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \dots$$

$$x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \dots$$

In the interval $[0, 2\pi]$:

The points are $\mathbf{x = \frac{\pi}{4}}$ and $\mathbf{x = \frac{5\pi}{4}}$.

9. Maximum value of $\sin x + \cos x$

We can rewrite the function:

$$f(x) = \sin x + \cos x = \sqrt{2} \left(\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x\right)$$

$$f(x) = \sqrt{2} (\cos \frac{\pi}{4} \sin x + \sin \frac{\pi}{4} \cos x) = \sqrt{2} \sin(x + \frac{\pi}{4})$$

Since the maximum value of $\sin(x + \pi/4)$ is 1,

  • Maximum Value: $\mathbf{\sqrt{2}}$.

10. Max/Min of $f(x) = 2x^3 – 24x + 107$

  1. $f'(x) = 6x^2 – 24 = 6(x^2 – 4)$. Critical points: $x = \pm 2$.

Case 1: Interval $[1, 3]$

  • Critical point in interval: $x = 2$.
  • Evaluate:
    • $f(1) = 2(1) – 24(1) + 107 = 85$
    • $f(2) = 2(8) – 24(2) + 107 = 16 – 48 + 107 = 75$
    • $f(3) = 2(27) – 24(3) + 107 = 54 – 72 + 107 = 89$
  • Maximum Value: $\mathbf{89}$.

Case 2: Interval $[-3, -1]$

  • Critical point in interval: $x = -2$.
  • Evaluate:
    • $f(-3) = 2(-27) – 24(-3) + 107 = -54 + 72 + 107 = 125$
    • $f(-2) = 2(-8) – 24(-2) + 107 = -16 + 48 + 107 = 139$
    • $f(-1) = 2(-1) – 24(-1) + 107 = -2 + 24 + 107 = 129$
  • Maximum Value: $\mathbf{139}$.

11. Find $a$ if $f(x) = x^4 – 62x^2 + ax + 9$ attains maximum at $x=1$ on $[0, 2]$

If a continuous function attains a local extremum in the interior of a closed interval, the derivative must be zero at that point.

  1. $f'(x) = 4x^3 – 124x + a$.
  2. Since the maximum occurs at $x=1$, an interior point, $f'(1) = 0$.$$f'(1) = 4(1)^3 – 124(1) + a = 0$$$$4 – 124 + a = 0 \implies -120 + a = 0$$$$\mathbf{a = 120}$$

12. Max/Min of $f(x) = x + \sin 2x$ on $[0, 2\pi]$

  1. $f'(x) = 1 + 2 \cos 2x$. Critical points: $\cos 2x = -1/2$.$$2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$$$$x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$$
  2. Evaluate: (Approximate values for comparison)
    • $f(0) = 0 + 0 = 0$
    • $f(\pi/3) = \pi/3 + \sin(2\pi/3) = \pi/3 + \sqrt{3}/2 \approx 1.047 + 0.866 \approx 1.913$
    • $f(2\pi/3) = 2\pi/3 + \sin(4\pi/3) = 2\pi/3 – \sqrt{3}/2 \approx 2.094 – 0.866 \approx 1.228$
    • $f(\pi) = \pi + \sin(2\pi) = \pi \approx 3.141$
    • $f(4\pi/3) = 4\pi/3 + \sin(8\pi/3) = 4\pi/3 + \sqrt{3}/2 \approx 4.189 + 0.866 \approx 5.055$
    • $f(5\pi/3) = 5\pi/3 + \sin(10\pi/3) = 5\pi/3 – \sqrt{3}/2 \approx 5.236 – 0.866 \approx 4.37$
    • $f(2\pi) = 2\pi + \sin(4\pi) = 2\pi \approx 6.28$
  3. Maximum Value: $\mathbf{2\pi}$ at $x=2\pi$.
  4. Minimum Value: $\mathbf{0}$ at $x=0$.

13-16. Numerical Optimization (Two Variables)

13. Two numbers sum to 24, product is maximum

Let the numbers be $x$ and $y$. Given: $x + y = 24 \implies y = 24 – x$.

Maximize $P = xy = x(24 – x) = 24x – x^2$.

  1. $P'(x) = 24 – 2x$. Critical point: $24 – 2x = 0 \implies x = 12$.
  2. $P”(x) = -2 < 0$, which is a maximum.
  3. The numbers are $x=12$ and $y = 24 – 12 = 12$.The numbers are $\mathbf{12}$ and $\mathbf{12}$.

14. Two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum

Given: $x + y = 60 \implies x = 60 – y$.

Maximize $P = xy^3 = (60 – y)y^3 = 60y^3 – y^4$.

  1. $P'(y) = 180y^2 – 4y^3 = 4y^2(45 – y)$.
  2. Critical points: $4y^2 = 0 \implies y = 0$ (min) or $45 – y = 0 \implies y = 45$.
  3. $y=45$: $x = 60 – 45 = 15$.
  4. $P”(y) = 360y – 12y^2$. $P”(45) = 360(45) – 12(45)^2 < 0$. (Maximum)The numbers are $\mathbf{x=15}$ and $\mathbf{y=45}$.

15. Two positive numbers $x$ and $y$ such that $x + y = 35$ and $x^2 y^5$ is a maximum

Given: $x + y = 35 \implies x = 35 – y$.

Maximize $P = x^2 y^5 = (35 – y)^2 y^5$.

  1. $P'(y) = 2(35 – y)(-1) y^5 + (35 – y)^2 (5y^4)$ (Product Rule)$P'(y) = y^4 (35 – y) [-2y + 5(35 – y)]$$P'(y) = y^4 (35 – y) [-2y + 175 – 5y] = y^4 (35 – y) (175 – 7y)$
  2. Critical points: $y = 0, y = 35, 175 – 7y = 0 \implies y = 25$.
  3. Since $x, y$ are positive, we test $y=25$. $x = 35 – 25 = 10$.
  4. Checking sign change or $P”(25)$ confirms maximum.The numbers are $\mathbf{x=10}$ and $\mathbf{y=25}$.

16. Two positive numbers sum to 16, sum of whose cubes is minimum

Let the numbers be $x$ and $y$. Given: $x + y = 16 \implies y = 16 – x$.

Minimize $S = x^3 + y^3 = x^3 + (16 – x)^3$.

  1. $S'(x) = 3x^2 + 3(16 – x)^2 (-1) = 3x^2 – 3(16 – x)^2$.
  2. Critical points: $3x^2 = 3(16 – x)^2 \implies x^2 = 256 – 32x + x^2$$$0 = 256 – 32x \implies 32x = 256 \implies x = 8$$
  3. $S”(x) = 6x + 6(16 – x) = 6x + 96 – 6x = 96$.
  4. $S”(8) = 96 > 0$, which is a minimum.The numbers are $x=8$ and $y = 16 – 8 = 8$.The numbers are $\mathbf{8}$ and $\mathbf{8}$.

17-26. Geometry Optimization (Application)

17. Maximum volume of a box from a square sheet ($18 \text{ cm} \times 18 \text{ cm}$)

Let $x$ be the side of the square cut from each corner.

The dimensions of the box are: length $L = 18 – 2x$, width $W = 18 – 2x$, height $H = x$.

Volume: $V(x) = x(18 – 2x)^2$. Domain: $0 < x < 9$.

  1. $V'(x) = 1 \cdot (18 – 2x)^2 + x \cdot 2(18 – 2x)(-2) = (18 – 2x) [(18 – 2x) – 4x]$$V'(x) = (18 – 2x) (18 – 6x)$.
  2. Critical points: $18 – 2x = 0 \implies x = 9$ (not in domain) or $18 – 6x = 0 \implies x = 3$.
  3. $V”(x) = \frac{d}{dx}(324 – 108x + 12x^2) = -108 + 24x$.
  4. $V”(3) = -108 + 24(3) = -108 + 72 = -36 < 0$. (Maximum)The side of the square to be cut off should be $\mathbf{3 \text{ cm}}$.

18. Maximum volume of a box from a rectangular sheet ($45 \text{ cm} \times 24 \text{ cm}$)

Let $x$ be the side of the square cut from each corner.

Dimensions: $L = 45 – 2x$, $W = 24 – 2x$, $H = x$.

Volume: $V(x) = x(45 – 2x)(24 – 2x)$. Domain: $0 < x < 12$.

  1. $V(x) = x(1080 – 138x + 4x^2) = 4x^3 – 138x^2 + 1080x$.
  2. $V'(x) = 12x^2 – 276x + 1080 = 12(x^2 – 23x + 90) = 12(x – 18)(x – 5)$.
  3. Critical points: $x = 18$ (not in domain) or $x = 5$.
  4. $V”(x) = 24x – 276$. $V”(5) = 24(5) – 276 = 120 – 276 = -156 < 0$. (Maximum)The side of the square to be cut off should be $\mathbf{5 \text{ cm}}$.

19. Largest rectangle inscribed in a fixed circle

Let the circle have radius $R$. If a rectangle with sides $x$ and $y$ is inscribed, the diagonal is $2R$.

$$x^2 + y^2 = (2R)^2 = 4R^2$$

Area: $A = xy$. Maximize $A^2 = x^2 y^2$.

Let $Z = A^2 = x^2 (4R^2 – x^2) = 4R^2 x^2 – x^4$.

  1. $\frac{dZ}{dx} = 8R^2 x – 4x^3 = 4x (2R^2 – x^2)$.
  2. Critical point: $2R^2 – x^2 = 0 \implies x^2 = 2R^2 \implies x = \sqrt{2}R$.
  3. If $x^2 = 2R^2$, then $y^2 = 4R^2 – 2R^2 = 2R^2$. Thus $y = \sqrt{2}R$.Since $x = y$, the rectangle is a square.

20. Right circular cylinder of given surface area and maximum volume

Given surface area $S$ (constant): $S = 2\pi r h + 2\pi r^2$.

Volume: $V = \pi r^2 h$.

  1. Eliminate $h$: $h = \frac{S – 2\pi r^2}{2\pi r} = \frac{S}{2\pi r} – r$.
  2. Substitute into $V$:$$V(r) = \pi r^2 \left(\frac{S}{2\pi r} – r\right) = \frac{Sr}{2} – \pi r^3$$
  3. Differentiate w.r.t. $r$: $\frac{dV}{dr} = \frac{S}{2} – 3\pi r^2$.
  4. Critical point: $\frac{S}{2} = 3\pi r^2 \implies S = 6\pi r^2$.
  5. Check maximum: $\frac{d^2 V}{dr^2} = -6\pi r < 0$. (Maximum)
  6. Find $h$: Substitute $S = 6\pi r^2$ into the $h$ equation:$$h = \frac{6\pi r^2}{2\pi r} – r = 3r – r = 2r$$Since $h = 2r$, the height is equal to the diameter of the base.

21. Minimum surface area for a given volume ($V = 100 \text{ cm}^3$)

Given $V = \pi r^2 h = 100$. Surface Area: $S = 2\pi r h + 2\pi r^2$.

  1. Eliminate $h$: $h = \frac{100}{\pi r^2}$.
  2. Substitute into $S$:$$S(r) = 2\pi r \left(\frac{100}{\pi r^2}\right) + 2\pi r^2 = \frac{200}{r} + 2\pi r^2$$
  3. Differentiate w.r.t. $r$: $\frac{dS}{dr} = -\frac{200}{r^2} + 4\pi r$.
  4. Critical point: $4\pi r = \frac{200}{r^2} \implies 4\pi r^3 = 200 \implies r^3 = \frac{50}{\pi}$.$$r = \sqrt[3]{\frac{50}{\pi}}$$
  5. Check minimum: $\frac{d^2 S}{dr^2} = \frac{400}{r^3} + 4\pi$. Since $r > 0$, $\frac{d^2 S}{dr^2} > 0$. (Minimum)
  6. Find $h$:$$h = \frac{100}{\pi r^2} = \frac{100}{\pi (\frac{50}{\pi})^{2/3}} = \frac{2 \cdot 50}{\pi \cdot \frac{50^{2/3}}{\pi^{2/3}}} = 2 \cdot 50^{1/3} \pi^{-1/3} = 2 \sqrt[3]{\frac{50}{\pi}} = 2r$$The dimensions are $r = \sqrt[3]{\frac{50}{\pi}} \text{ cm}$ and $h = 2r = 2 \sqrt[3]{\frac{50}{\pi}} \text{ cm}$.

22. Minimum combined area of a square and a circle

Total wire length $L = 28 \text{ m}$.

Let $x$ be the length used for the square (side $s=x/4$).

Let $y$ be the length used for the circle (circumference $C=y=2\pi r$). $y = 28 – x$.

$$s = x/4 \implies A_{\text{sq}} = (x/4)^2 = x^2/16$$

$$r = y/(2\pi) = (28 – x)/(2\pi) \implies A_{\text{circ}} = \pi r^2 = \pi \left(\frac{28 – x}{2\pi}\right)^2 = \frac{(28 – x)^2}{4\pi}$$

Total Area: $A(x) = \frac{x^2}{16} + \frac{(28 – x)^2}{4\pi}$. Domain: $0 \le x \le 28$.

  1. $\frac{dA}{dx} = \frac{2x}{16} + \frac{2(28 – x)(-1)}{4\pi} = \frac{x}{8} – \frac{(28 – x)}{2\pi}$.
  2. Critical point: $\frac{x}{8} = \frac{28 – x}{2\pi} \implies 2\pi x = 8(28 – x) \implies \pi x = 4(28 – x)$$$\pi x = 112 – 4x \implies x(\pi + 4) = 112 \implies x = \frac{112}{\pi + 4}$$
  3. Check minimum: $\frac{d^2 A}{dx^2} = \frac{1}{8} – \frac{(-1)}{2\pi} = \frac{1}{8} + \frac{1}{2\pi} > 0$. (Minimum)
  4. Lengths:
    • Square piece length $x = \mathbf{\frac{112}{\pi + 4}} \text{ m}$.
    • Circle piece length $y = 28 – x = 28 – \frac{112}{\pi + 4} = \frac{28(\pi + 4) – 112}{\pi + 4} = \frac{28\pi + 112 – 112}{\pi + 4} = \mathbf{\frac{28\pi}{\pi + 4}} \text{ m}$.

27-29. Multiple Choice Questions

27. Point on $x^2 = 2y$ nearest to $(0, 5)$

Let $P(x, y)$ be a point on the curve. $y = x^2/2$.

Distance squared $D^2 = D = (x – 0)^2 + (y – 5)^2 = x^2 + (x^2/2 – 5)^2$.

  1. $\frac{dD}{dx} = 2x + 2(x^2/2 – 5)(x) = 2x + x(x^2 – 10) = 2x + x^3 – 10x = x^3 – 8x = x(x^2 – 8)$.
  2. Critical points: $x = 0, x = \pm \sqrt{8} = \pm 2\sqrt{2}$.
  3. Evaluate Distance (Minimizing $D$ is equivalent to minimizing $D^2$):
    • $x=0$: $y=0$. $D^2 = 0^2 + (0 – 5)^2 = 25$.
    • $x=\pm 2\sqrt{2}$: $y = (\pm 2\sqrt{2})^2/2 = 8/2 = 4$. $D^2 = 8 + (4 – 5)^2 = 8 + 1 = 9$.The minimum distance is $\sqrt{9} = 3$ at points $(\pm 2\sqrt{2}, 4)$.The correct answer is (A) $(2\sqrt{2}, 4)$.

28. Minimum value of $\frac{x^2 – x + 1}{x^2 + x + 1}$

Let $y = \frac{x^2 – x + 1}{x^2 + x + 1}$.

$y(x^2 + x + 1) = x^2 – x + 1$

$x^2(y – 1) + x(y + 1) + (y – 1) = 0$.

For real values of $x$, the discriminant must be non-negative: $D \ge 0$.

$$D = (y + 1)^2 – 4(y – 1)(y – 1) = (y + 1)^2 – 4(y – 1)^2 \ge 0$$

$$(y^2 + 2y + 1) – 4(y^2 – 2y + 1) \ge 0$$

$$-3y^2 + 10y – 3 \ge 0 \implies 3y^2 – 10y + 3 \le 0$$

The roots of $3y^2 – 10y + 3 = 0$ are $y = \frac{10 \pm \sqrt{100 – 36}}{6} = \frac{10 \pm 8}{6}$, so $y = 3$ and $y = 1/3$.

Since the quadratic is facing upwards, $3y^2 – 10y + 3 \le 0$ means $1/3 \le y \le 3$.

The minimum value is $\mathbf{1/3}$.

The correct answer is (D) $\frac{1}{3}$.

29. Maximum value of $\frac{1}{3}[x(x – 1) + 1]$, $0 \le x \le 1$

Let $f(x) = \frac{1}{3}(x^2 – x + 1)$.

  1. $f'(x) = \frac{1}{3}(2x – 1)$. Critical point: $2x – 1 = 0 \implies x = 1/2$.
  2. Evaluate:
    • $f(0) = \frac{1}{3}(0 – 0 + 1) = 1/3$
    • $f(1/2) = \frac{1}{3}(1/4 – 1/2 + 1) = \frac{1}{3}(3/4) = 1/4$
    • $f(1) = \frac{1}{3}(1 – 1 + 1) = 1/3$The maximum value is $\mathbf{1/3}$.The correct answer is (A) $\frac{1}{3}$.
Ex 6.1
Ex 6.2
Ex 6.3
Ch 6 Misc.

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