Exercise 6.3 Complete Guide

(i) $f(x) = (2x – 1)^2 + 3$ Since $(2x-1)^2 \ge 0$, min value occurs when $2x-1=0 \implies x=1/2$.
Min Value = 3. No Max Value. (ii) $f(x) = 9x^2 + 12x + 2$ $f(x) = (3x+2)^2 – 2$. Min occurs when $3x+2=0 \implies x=-2/3$.
Min Value = -2. No Max Value. (iii) $f(x) = -(x – 1)^2 + 10$ Max value occurs when $-(x-1)^2 = 0 \implies x=1$.
Max Value = 10. No Min Value. (iv) $g(x) = x^3 + 1$ $g'(x) = 3x^2 \ge 0$. Function is strictly increasing.
No Max or Min Value.

(i) $f(x) = |x + 2| – 1$ $|x+2| \ge 0$. Min at $x=-2$. Min Value = -1. No Max. (ii) $g(x) = -|x + 1| + 3$ Max value when $|x+1|=0$. Max Value = 3. No Min. (iii) $h(x) = \sin(2x) + 5$ Range of $\sin(2x)$ is $[-1, 1]$.
Min = $-1+5=4$. Max = $1+5=6$. (iv) $f(x) = |\sin 4x + 3|$ Range of $\sin 4x$ is $[-1, 1]$. Term inside mod is $[2, 4]$.
Min = 2. Max = 4. (v) $h(x) = x + 1, x \in (-1, 1)$ Open interval has no endpoints. No Max or Min.

(i) $f(x) = x^2$ $f’=2x=0 \implies x=0$. $f”=2 > 0$ (Min). Local Min = 0. (ii) $g(x) = x^3 – 3x$ $g’=3x^2-3=0 \implies x=\pm 1$. $g”=6x$.
$x=1 \to$ Min (-2). $x=-1 \to$ Max (2). (iii) $h(x) = \sin x + \cos x, 0 < x < \pi/2$ $h’=\cos x – \sin x = 0 \implies x=\pi/4$. $h” < 0$. Max = $\sqrt{2}$. (iv) $f(x) = \sin x – \cos x, 0 < x < 2\pi$ $f’=\cos x + \sin x = 0 \implies \tan x = -1 \implies x = 3\pi/4, 7\pi/4$.
$x=3\pi/4 \to$ Max ($\sqrt{2}$). $x=7\pi/4 \to$ Min ($-\sqrt{2}$). (v) $f(x) = x^3 – 6x^2 + 9x + 15$ $f’=3(x^2-4x+3)=3(x-1)(x-3)=0$.
$x=1 \to$ Max (19). $x=3 \to$ Min (15). (vi) $g(x) = x/2 + 2/x, x > 0$ $g’=1/2 – 2/x^2=0 \implies x=2$. Min at $x=2$. Value = 2. (vii) $g(x) = 1/(x^2+2)$ $g’ = -2x/(x^2+2)^2$. $x=0$. Max at $x=0$. Value = 1/2. (viii) $f(x) = x\sqrt{1-x}, x < 1$ $f’ = \frac{2-3x}{2\sqrt{1-x}}$. Crit point $x=2/3$. Max = $2/(3\sqrt{3})$.

$f'(x) = e^x$. Since $e^x > 0$ for all real $x$, the slope is always positive. The function is strictly increasing. Thus, no maxima or minima exists.

(i) $f(x)=x^3, x \in [-2, 2]$ $f’=3x^2=0 \implies x=0$. Check endpoints and 0.
$f(-2)=-8, f(0)=0, f(2)=8$. Abs Max=8, Abs Min=-8. (ii) $f(x)=\sin x + \cos x, x \in [0, \pi]$ $f’=0 \implies x=\pi/4$.
$f(0)=1, f(\pi/4)=\sqrt{2}, f(\pi)=-1$. Abs Max=$\sqrt{2}$, Abs Min=-1. (iii) $f(x)=4x – \frac{1}{2}x^2, x \in [-2, 4.5]$ $f’=4-x=0 \implies x=4$.
$f(-2)=-10, f(4)=8, f(4.5)=7.875$. Abs Max=8, Abs Min=-10. (iv) $f(x)=(x-1)^2 + 3, x \in [-3, 1]$ Min at vertex $x=1$ (Value 3). Max at $x=-3$ (Value 19).

$p'(x) = -72 – 36x$. Set $p'(x)=0 \implies 36x = -72 \implies x = -2$.
$p”(-2) = -36 < 0$ (Max).
Max Profit $= 41 – 72(-2) – 18(4) = 41 + 144 – 72 = 113$.

$f'(x) = 12x^3 – 24x^2 + 24x – 48 = 12(x-2)(x^2+2)$.
Real critical point: $x=2$.
$f(0) = 25$.
$f(2) = 48 – 64 + 48 – 96 + 25 = -39$.
$f(3) = 243 – 216 + 108 – 144 + 25 = 16$.

$f'(x) = 2\cos(2x) = 0 \implies 2x = \pi/2, 3\pi/2, 5\pi/2, 7\pi/2$.
$x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$.
Values: $1, -1, 1, -1$.

Let $f(x) = \sin x + \cos x$. $f'(x) = \cos x – \sin x = 0 \implies \tan x = 1 \implies x = \pi/4$.
Max Value $= \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.

$f'(x) = 6x^2 – 24 = 6(x^2 – 4)$. Critical points $x = \pm 2$.
Interval $[1, 3]$: Use $x=2$. $f(1)=85, f(2)=75, f(3)=89$. Max=89.
Interval $[-3, -1]$: Use $x=-2$. $f(-3)=125, f(-2)=139, f(-1)=129$. Max=139.

$f'(x) = 4x^3 – 124x + a$.
At $x=1$, $f'(1) = 0$.
$4(1) – 124(1) + a = 0 \implies a = 120$.

$f'(x) = 1 + 2\cos 2x = 0 \implies \cos 2x = -1/2$.
$2x = 2\pi/3, 4\pi/3, 8\pi/3, 10\pi/3$.
$x = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$.
Checking values: $f(0)=0, f(2\pi)=2\pi$.
Max is at $2\pi$, Min is at $0$.

Let numbers be $x$ and $24-x$.
$P(x) = x(24-x) = 24x – x^2$.
$P'(x) = 24 – 2x = 0 \implies x = 12$.

$x = 60-y$. Let $f(y) = (60-y)y^3 = 60y^3 – y^4$.
$f'(y) = 180y^2 – 4y^3 = 4y^2(45-y)$.
Crit point $y=45$. Then $x=15$.

$x = 35-y$. $f(y) = (35-y)^2 y^5$.
$f'(y) = y^5 \cdot 2(35-y)(-1) + (35-y)^2 \cdot 5y^4$.
$f'(y) = y^4(35-y)[-2y + 5(35-y)] = y^4(35-y)(175 – 7y)$.
$175 – 7y = 0 \implies y = 25$. Then $x = 10$.

$S = x^3 + (16-x)^3$.
$S’ = 3x^2 + 3(16-x)^2(-1) = 3[x^2 – (256 + x^2 – 32x)] = 3(32x – 256)$.
$32x – 256 = 0 \implies x = 8$.

Side of cut square = $x$.
$V = x(18-2x)^2$.
$V’ = (18-2x)^2 + x \cdot 2(18-2x)(-2) = (18-2x)(18-6x)$.
$18-6x=0 \implies x=3$.

$V = x(45-2x)(24-2x) = 4x(45-2x)(12-x)$.
$V'(x) = 12(x-5)(x-18) = 0$.
$x=18$ is not possible (width is 24). So $x=5$.

Circle radius $R$ (constant). Rectangle sides $x, y$.
$x^2 + y^2 = (2R)^2 = 4R^2 \implies y = \sqrt{4R^2-x^2}$.
$A = xy$. Maximize $A^2 = x^2(4R^2-x^2) = 4R^2x^2 – x^4$.
Diff: $8R^2x – 4x^3 = 0 \implies x^2 = 2R^2 \implies x = \sqrt{2}R$.
Then $y = \sqrt{4R^2 – 2R^2} = \sqrt{2}R$.
Since $x=y$, it is a square.

$S = 2\pi r^2 + 2\pi rh \implies h = \frac{S – 2\pi r^2}{2\pi r}$.
$V = \pi r^2 h = \pi r^2 \left(\frac{S – 2\pi r^2}{2\pi r}\right) = \frac{1}{2}(Sr – 2\pi r^3)$.
$V’ = \frac{1}{2}(S – 6\pi r^2) = 0 \implies S = 6\pi r^2$.
Substitute $S$: $2\pi r^2 + 2\pi rh = 6\pi r^2 \implies 2\pi rh = 4\pi r^2 \implies h = 2r$.

$V = \pi r^2 h = 100 \implies h = 100/(\pi r^2)$.
$S = 2\pi r^2 + 2\pi r(100/\pi r^2) = 2\pi r^2 + 200/r$.
$S’ = 4\pi r – 200/r^2 = 0 \implies \pi r^3 = 50$.
Also implies $h = 2r$.

Let cut be $x$ (circle) and $28-x$ (square).
Radius $r = x/2\pi$, Side $a = (28-x)/4$.
$A = \pi (x/2\pi)^2 + ((28-x)/4)^2$.
$dA/dx = x/2\pi – (28-x)/8 = 0 \implies 4x = \pi(28-x)$.
$x(4+\pi) = 28\pi \implies x = \frac{28\pi}{4+\pi}$.

Let Sphere Radius $R$, Cone height $h$, Cone radius $r$. Let $x$ be distance center to base.
$h = R+x, r^2 = R^2-x^2$.
$V = \frac{1}{3}\pi(R^2-x^2)(R+x)$.
$V’ = 0 \implies x = R/3$.
$h = 4R/3, r^2 = 8R^2/9$.
$V_{max} = \frac{1}{3}\pi (8R^2/9)(4R/3) = \frac{32}{81}\pi R^3$.
Sphere Vol $V_s = \frac{4}{3}\pi R^3$.
Ratio: $\frac{32\pi R^3/81}{4\pi R^3/3} = \frac{8}{27}$.

Minimize $S = \pi r l$ subject to $V = \frac{1}{3}\pi r^2 h$.
Work with $S^2$ and substitute $l^2 = r^2+h^2$.
Result: Height must be $\sqrt{2}$ times the radius ($h = \sqrt{2}r$).

$V = \frac{1}{3}\pi r^2 h$. Use $r^2 = l^2 – h^2$.
$V(h) = \frac{\pi}{3}(l^2 h – h^3)$.
$V’ = \frac{\pi}{3}(l^2 – 3h^2) = 0 \implies l^2 = 3h^2 \implies l = \sqrt{3}h$.
$\cos \alpha = h/l = 1/\sqrt{3}$.
$\tan \alpha = \sqrt{\sec^2\alpha – 1} = \sqrt{3-1} = \sqrt{2}$.
$\alpha = \tan^{-1}\sqrt{2}$.

Minimize $S = \pi r l + \pi r^2$.
Lengthy derivation involving setting $dS/dr = 0$ yields $l = 3r$.
$\sin \alpha = r/l = r/3r = 1/3$.
$\alpha = \sin^{-1}(1/3)$.

Minimize distance square $D = x^2 + (y-5)^2 = 2y + (y-5)^2$.
$D’ = 2 + 2(y-5) = 2y – 8 = 0 \implies y=4$.
$x^2 = 2(4) = 8 \implies x = \pm 2\sqrt{2}$.
Wait, re-evaluating options from NCERT.
Standard Solution: $(2\sqrt{2}, 4)$.
Correct Answer: Option (A)

Let $y = \frac{1-x+x^2}{1+x+x^2}$. $y’ = \frac{2(x^2-1)}{(1+x+x^2)^2} = 0 \implies x = \pm 1$.
$x=1 \implies y = 1/3$ (Min).
$x=-1 \implies y = 3$ (Max).
Correct Answer: Option (D) 1/3

Let $g(x) = x^2 – x + 1$. Vertex at $x=1/2$, value $3/4$.
Endpoints $g(0)=1, g(1)=1$.
Max of inner function is 1. Max of total function is $1^{1/3} = 1$.
Correct Answer: Option (C) 1