Optimization

Find two nonnegative numbers whose sum is 50 such that their product is as large as possible.

Solution

Let one number be $x$; the other is $50-x$. Their product is $$f(x)=x(50-x)=50x-x^2.$$ Because both numbers must be nonnegative, the domain is $[0,50]$. Since $f$ is continuous on the closed interval $[0,50]$, it attains an absolute maximum. The only critical point comes from f'(x)=50-2x=0\Rightarrow x=25. Comparing values at the critical point and endpoints: $$f(0)=0,f(25)=625,f(50)=0.$$ The maximum product is $\mathbf{625}$, achieved when both numbers equal $25$.

Find the dimensions of the rectangle of greatest area that can be inscribed in a circle of radius 2.

Solution

Inscribe a rectangle $BCDE$ in the circle. Let $x=CD$. By the Pythagorean theorem, $DE=\sqrt{4^2-x^2}$, so the area is $$A(x)=x\sqrt{16-x^2},0\le x\le 4.$$

Setting A'(x)=0: A'(x)=\sqrt{16-x^{2}}-\frac{x^{2}}{\sqrt{16-x^{2}}}=0\Rightarrow16-2x^{2}=0\Rightarrow x=2\sqrt{2}. Comparing values: $$A(0)=0,A(2\sqrt{2})=\sqrt{8}\cdot \sqrt{8}=8,A(4)=0.$$ The rectangle of maximum area is a **square** with side lengths $2\sqrt{2}$ and area $8$.

A piece of cardboard measures $48\text{ in}\times 18\text{ in}$. An open box is made by cutting equal squares of side $x$ from each corner and folding up the sides. Find the cut size $x$ that maximizes the volume.

Solution

After cutting, the box has dimensions $48-2x$, $18-2x$, and $x$ (height):

$$V(x)=(48-2x)(18-2x)x=4x^3-132x^2+864x.$$ The dimensions must be nonnegative, so $0\le x\le 9$. Setting V'(x)=0: V'(x)=12x^{2}-264x+864=12(x^{2}-22x+72)=12(x-4)(x-18)=0 $$\Rightarrow x=4\text{ or }x=18.$$ Only $x=4\in [0,9]$. Comparing values: | $x$ | $0$ | $4$ | $9$ | |-----|-----|------|-----| | $V(x)$ | $0$ | $1600$ | $0$ | The maximum volume is $\mathbf{1600}{\mathbf{\text{in}}}^{\mathbf{3}}$, obtained when $x=4$ in.

What is the shortest distance from the point $(0,2)$ to the parabola $8-x^2=2y$?

Solution

Any point on the parabola has the form $(x,\frac{1}{2}(8-x^2))$. The distance from $(0,2)$ to this point is $$D=\sqrt{x^2+{(\frac{1}{2}(8-x^2)-2)}^2}=\sqrt{x^2+{(2-\frac{1}{2}{x}^2)}^2}.$$ To avoid differentiating a square root, minimize $D^2$: $$f(x)=x^2+{(2-\frac{1}{2}{x}^2)}^2=\frac{1}{4}{x}^4-x^2+4.$$ Note: $f$ is even, so we may analyze $[0,\mathrm{\infty })$. Setting f'(x)=0: f'(x)=x^{3}-2x=x(x^{2}-2)=0\Rightarrow x=0,\pm\sqrt{2}. Sign diagram for f':

By the First Derivative Test, $x=0$ gives a **local maximum** and $x=\pm \sqrt{2}$ give **local minima**. (The Second Derivative Test fails at $x=\pm \sqrt{2}$ since f''(\pm\sqrt{2})=0.) The minimum distance is $$D{|}_{x=\pm \sqrt{2}}=\sqrt{2+{(2-1)}^2}=\sqrt{3}.$$ The closest points on the parabola to $(0,2)$ are $(\sqrt{2},3)$ and $(-\sqrt{2},3)$.

Find the height and radius of the right circular cylinder of maximum volume that can be inscribed in a right circular cone with radius $18\text{ cm}$ and height $30\text{ cm}$.

Solution

Let $r$ = radius of the cylinder, $h$ = height of the cylinder, $V$ = volume. **Primary equation:** $V=\pi r^{2}h$. **Secondary equation (similar triangles):** Looking at the cone from the side, triangles $OAB$ and O'AB' are similar:

$$\frac{30-h}{30}=\frac{r}{18}\Rightarrow h=30-\frac{5}{3}r.$$ Substituting: $$V(r)=\pi r^2(30-\frac{5}{3}r)=30\pi r^2-\frac{5}{3}\pi r^3,0\le r\le 18.$$ Setting V'(r)=0: V'(r)=60\pi r-5\pi r^{2}=5\pi r(12-r)=0\Rightarrow r=0\text{ or }r=12. Comparing values: | $r$ | $0$ | $12$ | $18$ | |-----|-----|------|------| | $V$ | $0$ | $1440\pi$ | $0$ | Maximum volume occurs at $r=12\text{ cm}$, $h=30-\frac{5}{3}(12)=10\text{ cm}$.

The lower corner of a long page of width $a$ is folded over so as just to reach the inner edge of the page.

(a) Find the width $x$ of the folded part when the length of the crease is a minimum.

(b) Find the width when the area folded over is a minimum.

Solution

Let $x=$ length of $BC$, $y=$ length of $CE$, $L=$ length of the crease $BE$.

From $\mathrm{△}BCE$: $L^2=x^2+y^2$. Using the geometry of the fold, the constraint between $x$ and $y$ works out to y^{2}=\frac{ax^{2}}{2x-a}.\tag{ii} **(a) Minimize the crease length $L$.** Substituting (ii) and simplifying: Setting f'(x)=0: f'(x)=\frac{8x^{3}-6ax^{2}}{(2x-a)^{2}}=\frac{2x^{2}(4x-3a)}{(2x-a)^{2}}=0\Rightarrow x=\frac{3a}{4}. Since f'>0 for and f'<0 for , by the First Derivative Test $x=3a/4$ is a **local minimum**. Since $L\to +\mathrm{\infty }$ as $x\to (a/2{)}^{+}$ and as $x\to +\mathrm{\infty }$, this local minimum is also the absolute minimum. The minimum crease length is $$L=\sqrt{\frac{2(3a/4{)}^3}{2(3a/4)-a}}=\frac{3\sqrt{3}}{4}a\approx 1.299a.$$

**(b) Minimize the folded area $A=xy$.** Substituting (ii): Minimize $g(x)={\displaystyle \frac{a{x}^4}{2x-a}}$. Setting g'(x)=0: g'(x)=\frac{4ax^{3}(2x-a)-2ax^{4}}{(2x-a)^{2}}=\frac{2ax^{3}(4x-3a-x)}{(2x-a)^{2}}=0\Rightarrow x=\frac{2a}{3}. Check: $x=2a/3\in (a/2,a]$. The minimum folded area is $$A{|}_{x=2a/3}=\frac{4a^2}{3\sqrt{3}}.$$

Two corridors of widths $a$ and $b$ meet at right angles. What is the maximum length of a ladder that can be carried horizontally around the corner?

Solution

**Method (a): Cartesian coordinates.**

The total length of the ladder is $L=L_1+L_2=\sqrt{x^2+b^2}+\sqrt{y^2+a^2}$. From the similar-triangle constraint $y=ab/x$: $$L=\sqrt{x^2+b^2}+\frac{a}{x}\sqrt{x^2+b^2}=\sqrt{x^2+b^2}(1+\frac{a}{x}).$$ Setting $dL/dx=0$: $$\frac{dL}{dx}=0\Rightarrow x-\frac{a{b}^2}{x^2}=0\Rightarrow x=\sqrt[3]{a{b}^2}.$$ At this critical point: $$L={(a^{2/3}+b^{2/3})}^{3/2}.$$ Since $L\to +\mathrm{\infty }$ as $x\to 0$ or $x\to +\mathrm{\infty }$, the local minimum is the absolute minimum. **Method (b): Trigonometric parametrization.** Setting L'(\theta)=0: $$-\frac{b\cos \theta }{{\sin }^2\theta }+\frac{a\sin \theta }{{\cos }^2\theta }=0\Rightarrow {\tan }^3\theta =\frac{b}{a}\Rightarrow \theta =\arctan \sqrt[3]{\frac{b}{a}}.$$ Using $\sin (\arctan \alpha )={\displaystyle \frac{\alpha }{\sqrt{1+{\alpha }^2}}}$ and $\cos (\arctan \alpha )={\displaystyle \frac{1}{\sqrt{1+{\alpha }^2}}}$:

$$L\left(\arctan \sqrt[3]{\frac{b}{a}}\right)={(a^{2/3}+b^{2/3})}^{3/2}.$$ Both methods confirm: the longest ladder that can be carried around the corner has length $𝑳\mathbf{=}{({𝒂}^{\mathbf{2}\mathbf{/}\mathbf{3}}\mathbf{+}{𝒃}^{\mathbf{2}\mathbf{/}\mathbf{3}})}^{\mathbf{3}\mathbf{/}\mathbf{2}}$.