Optimization (Maxima/Minima)

{Velocity of a Moving Particle} The position of a particle is given by $s(t) = 5t^2 + 2t$. Find its velocity at $t = 3$ seconds.

Solution: Velocity is the derivative of position: \[ v(t) = s'(t) = \frac{d}{dt}(5t^2 + 2t) = 10t + 2 \] At $t=3$: \[ v(3) = 10(3) + 2 = 32 \text{ units/sec} \]

{Marginal Cost} The cost (in dollars) of producing $x$ units of a product is $C(x) = 50 + 4x + 0.1x^2$. Find the marginal cost when $x=20$ units.

Solution: Marginal cost is the derivative of the total cost: \[ C'(x) = \frac{d}{dx} (50 + 4x + 0.1x^2) = 4 + 0.2x \] At $x=20$: \[ C'(20) = 4 + 0.2(20) = 4 + 4 = 8 \text{ dollars/unit} \]

{Rate of Change of Temperature} The temperature of a cup of coffee at time $t$ minutes is modeled by $T(t) = 80 - 5\ln(t+1)$. Find the rate at which the temperature is decreasing at $t=4$ minutes.

Solution: Rate of change of temperature: \[ T'(t) = \frac{d}{dt} \left(80 - 5\ln(t+1)\right) = -\frac{5}{t+1} \] At $t=4$: \[ T'(4) = -\frac{5}{5} = -1 \text{ degree/minute} \] Conclusion: The coffee is cooling at 1 degree per minute at $t=4$.

{Rate of Change of Area} The side of a square grows with time according to $s(t) = 3t + 2$ meters. Find the rate at which the area of the square changes at $t=2$ seconds.

Solution: Area $A = s^2$, so \[ \frac{dA}{dt} = 2s \cdot \frac{ds}{dt} = 2(3t+2) \cdot 3 = 6(3t+2) \] At $t=2$: \[ \frac{dA}{dt} = 6(6+2) = 6 \cdot 8 = 48 \text{ m}^2/\text{sec} \]

Determine all critical points of \[ f(x) = 4x^3 - 12x^2 + 7x + 5 \]

Solution: Derivative: \[ f'(x) = 12x^2 - 24x + 7 \] Set $f'(x) = 0$: \[ 12x^2 - 24x + 7 = 0 \implies x = 1 \pm \tfrac{\sqrt{15}}{6} \]

Conclusion: Critical points are $x = 1 \pm \tfrac{\sqrt{15}}{6}$.

Find the critical points of \[ g(t) = \sqrt{t}(t-2), \quad t \geq 0 \]

Solution: Derivative: \[ g'(t) = \frac{1}{2\sqrt{t}}(t-2) + \sqrt{t} = \frac{3t-2}{2\sqrt{t}} \]

Set $g'(t)=0$: \[ 3t-2=0 \implies t=\tfrac{2}{3} \]

Also, $g'(t)$ undefined at $t=0$ (but $g(0)=0$ exists).

Conclusion: Critical points are $t=0$ and $t=\tfrac{2}{3}$.

Find the critical points of \[ y = 5x - 3\sin(2x) \]

Solution: Derivative: \[ y' = 5 - 6\cos(2x) \]

Set $y'=0$: \[ \cos(2x)=\tfrac{5}{6} \]

Conclusion: Infinite critical points: \[ x=\tfrac{1}{2}\cos^{-1}\!\left(\tfrac{5}{6}\right)+n\pi,\quad n\in\mathbb{Z} \]

Determine all critical points of \[ h(t) = t e^{-t^2} \]

Solution: Derivative: \[ h'(t)=e^{-t^2}(1-2t^2) \]

Set $h'(t)=0$: \[ 1-2t^2=0 \implies t=\pm \tfrac{1}{\sqrt{2}} \]

Conclusion: Critical points at $t=\pm \tfrac{1}{\sqrt{2}}$.

Find critical points of \[ f(x)=x^2 \ln(x),\quad x>0 \]

Solution: Derivative: \[ f'(x)=2x\ln x + x = x(2\ln x+1) \]

Set $f'(x)=0$: $x=0$ (not in domain), or $\ln x=-\tfrac{1}{2} \implies x=e^{-1/2}$.

Conclusion: Critical point at $x=e^{-1/2}$.

Determine all critical points of \[ f(x)=x e^{x^2} \]

Solution: Derivative: \[ f'(x)=e^{x^2}(1+2x^2) \]

Since $e^{x^2}>0$ for all $x$, we need $1+2x^2=0$, which has no real solution.

Conclusion: No critical points in the real domain.

{Quadratic Function on a Closed Interval} Find the absolute and relative extrema of \[ f(x) = x^2 - 4x + 3 \quad \text{on } [0,3] \] Solution:
Derivative: $f'(x) = 2x - 4$. Set $f'(x)=0 \implies 2x-4=0 \implies x=2$ (critical point).

Check endpoints and critical point: \[ f(0)=3, \quad f(2) = -1, \quad f(3)=0 \]

Conclusion: Absolute maximum = $f(0)=3$, Absolute minimum = $f(2)=-1$, Relative minimum = $f(2)=-1$, Relative maximum = $f(0)=3$ (endpoint, only absolute).

{Cubic Function on a Closed Interval} Find absolute and relative extrema of \[ f(x) = x^3 - 3x^2 + 2 \quad \text{on } [-1,2] \] Solution:
Derivative: $f'(x) = 3x^2 - 6x = 3x(x-2)$. Critical points: $x=0,2$

Check endpoints and critical points: \[ f(-1) = -2, \quad f(0)=2, \quad f(2)= -2 \]

Conclusion: Absolute maximum = $f(0)=2$, Absolute minimum = $f(-1) = -2$ (also occurs at $x=2$), Relative maximum = $f(0)=2$, Relative minimum = none in interior.

{Trigonometric Function} Find extrema of \[ f(x) = \sin x \quad \text{on } [0,2\pi] \] Solution:
Derivative: $f'(x) = \cos x$. Set $f'(x)=0 \implies \cos x = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}$

Check endpoints and critical points: \[ f(0)=0, f(2\pi)=0, f(\pi/2)=1, f(3\pi/2)=-1 \]

Conclusion: Absolute maximum = 1 at $x=\pi/2$, Absolute minimum = -1 at $x=3\pi/2$, Relative maximum = 1 at $x=\pi/2$, Relative minimum = -1 at $x=3\pi/2$.

{Function with No Relative Extrema} \[ f(x) = x^3 \quad \text{on } [-2,2] \] Solution:
Derivative: $f'(x)=3x^2=0 \implies x=0$

Check second derivative or graph: $f”(x)=6x$ At $x=0$, $f”(0)=0$ → inflection point, not maximum or minimum

Check endpoints for absolute extrema: $f(-2)=-8, f(2)=8$

Conclusion: Absolute maximum = 8 at $x=2$, Absolute minimum = -8 at $x=-2$, No relative extrema.

To find the absolute extrema of $f(x)$ on $[a,b]$:

1. Verify that $f(x)$ is continuous on $[a,b]$.

2. Find all critical points $c$ in $(a,b)$ where $f'(c)=0$ or $f'(c)$ does not exist.

3. Evaluate $f(x)$ at all critical points and at the endpoints $x=a$ and $x=b$.

4. Identify the largest and smallest values; these are the absolute maximum and absolute minimum, respectively.

{Polynomial Function on a Closed Interval} Find the absolute extrema of \[ f(x) = x^3 - 6x^2 + 9x + 1 \quad \text{on } [0,4] \] Solution:
Derivative: $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$

Critical points in $[0,4]$: $x=1,3$

Evaluate $f(x)$ at endpoints and critical points: \[ f(0)=1, \quad f(1)=5, \quad f(3)=1, \quad f(4)=1 \]

Conclusion: Absolute maximum = $f(1)=5$, Absolute minimum = $f(0)=f(3)=f(4)=1$

{Cubic Function with Restricted Interval} Find the absolute extrema of \[ f(x) = 2x^3 - 9x^2 + 12x + 5 \quad \text{on } [1,3] \] Solution:
Derivative: $f'(x) = 6x^2 - 18x + 12 = 6(x^2 -3x + 2) = 6(x-1)(x-2)$

Critical points in $[1,3]$: $x=1,2$ (ignore $x=1$ if already endpoint)

Evaluate $f(x)$ at endpoints and critical points: \[ f(1)=10, \quad f(2)=8, \quad f(3)=11 \]

Conclusion: Absolute maximum = $f(3)=11$, Absolute minimum = $f(2)=8$

{Trigonometric Function} Determine the minimum and maximum values of \[ f(t) = \sin(2t) + t \quad \text{on } [0, \pi] \] Solution:
Derivative: $f'(t) = 2\cos(2t) + 1$

Set $f'(t)=0 \implies 2\cos(2t)+1=0 \implies \cos(2t)=-\frac{1}{2} \implies 2t = \frac{2\pi}{3}, \frac{4\pi}{3} \implies t = \frac{\pi}{3}, \frac{2\pi}{3}$

Evaluate at critical points and endpoints: \[ f(0)=0, \quad f(\pi)=\pi, \quad f(\pi/3) = \frac{\sqrt{3}}{2} + \frac{\pi}{3}, \quad f(2\pi/3) = -\frac{\sqrt{3}}{2} + \frac{2\pi}{3} \]

Conclusion: Absolute maximum = $f(\pi) = \pi$, Absolute minimum = $f(0)=0$

{Exponential Function} Find the extrema of \[ f(x) = x e^{-x^2} \quad \text{on } [-2,2] \] Solution:
Derivative: $f'(x) = e^{-x^2} - 2x^2 e^{-x^2} = e^{-x^2}(1-2x^2)$

Critical points: $1-2x^2=0 \implies x=\pm \frac{1}{\sqrt{2}}$

Evaluate at endpoints and critical points: \[ f(-2) \approx -0.036, \quad f(2)\approx 0.036, \quad f(\pm 1/\sqrt{2}) = \pm 0.43 \]

Conclusion: Absolute maximum $\approx 0.43$ at $x=1/\sqrt{2}$, Absolute minimum $\approx -0.43$ at $x=-1/\sqrt{2}$

{Function with Nonexistent Derivative at Critical Point} Determine extrema of \[ f(x) = |x-1| \quad \text{on } [0,3] \] Solution:
Derivative does not exist at $x=1$ (critical point).

Evaluate endpoints and critical point: \[ f(0)=1, \quad f(1)=0, \quad f(3)=2 \]

Conclusion: Absolute maximum = $f(3)=2$, Absolute minimum = $f(1)=0$

Let $f(x) = x^3 - 3x^2 + 2$.
Compute derivative: $f'(x) = 3x^2 - 6x = 3x(x-2)$.
Critical points: $x=0,2$.
Test intervals: - $(-\infty,0)$: $f'(x)>0$ → increasing - $(0,2)$: $f'(x)<0$ → decreasing - $(2,\infty)$: $f'(x)>0$ → increasing

Let $f(x) = x^3 - 3x^2 + 2$.
Second derivative: $f”(x) = 6x - 6$.
Set $f”(x)=0 \implies x=1$.
Test intervals: - $(-\infty,1)$: $f”(x)<0$ → concave down - $(1,\infty)$: $f”(x)>0$ → concave up. Inflection point at $x=1$.