6.13 Optimization

If one of those numbers is $x$, the other one is $50-x$. Therefore, their product is $$x(50-x)$$ and because $x$ and $50-x$ are nonnegative numbers, we have $0\le x\le 50$ (otherwise $x$ or $50-x$ will be negative). That is, we want to maximize $$\begin{array}{}\text{(*)}& f(x)=x(50-x)=50x-x^2\end{array}$$ on the interval $[0,50]$. Because $f$ is a continuous function and the interval is closed, $f$ attains its absolute maximum (and its absolute minimum) on this interval, and its extreme values occur either at a critical point or an endpoint. Because $f$ is a polynomial and hence is differentiable everywhere, its critical points are only the points at which the derivative vanishes $f^{\prime }(x)=0$. From (*) we obtain $$f^{\prime }(x)=50-2x=0\iff x=25.$$ Therefore, the absolute maximum of $f$ is one of the following numbers $$f(0)=0,f(25)=25(50-25)=625,f(50)=0.$$ Thus, the maximum value of $f$ is 625, which occurs at $x=25$, and hence the two nonnegative numbers are $x=25$ and $50-x=25$.

The graph of $f$ is shown in Figure 1.

Inscribe any rectangle, as $BCDE$ (Figure 2). Let $$x=CD$$ then $DE=\sqrt{4^2-x^2}$, and the area of the rectangle is $$A(x)=CD\cdot DE=x\sqrt{16-x^2}.$$

 

Because $0\le x\le 4$, the problem reduces to maximization of $A(x)$ on the interval $[0,4]$. $$A^{\prime }(x)=\sqrt{16-x^2}+x\frac{-2x}{2\sqrt{16-x^2}}.$$ Setting $A^{\prime }(x)=0$, we obtain $$\sqrt{16-x^2}-\frac{x^2}{\sqrt{16-x^2}}=0.$$ Multiplying both sides by $\sqrt{16-x^2}$: $$16-x^2-x^2=0$$ $$\Rightarrow x=\pm \sqrt{8}=\pm 2\sqrt{2}.$$ Only $x=2\sqrt{2}$ is in the interval $[0,4]$. Therefore, the absolute maximum of $A$ is one of the following numbers: $$A(0)=0,A(4)=0,$$ and $$A(2\sqrt{2})=\sqrt{8}\sqrt{16-8}=8.$$ The above calculations show us that the rectangle of maximum area inscribed in the circle is a square with side lengths of $\sqrt{8}=2\sqrt{2}$ and of area 8.

The graph of $A$ is shown in Figure 3.

Let
$x=$ length of each side of the square
$V=$ volume of the open box
We need to find a relationship between $V$ and $x$. Because the dimensions of the box are $48-2x$, $18-2x$, and $x$ (Figure 6), the volume of the box is $$V=(48-2x)(18-2x)x=4x^3-132x^2+864x.$$

Because the dimensions of the box must be nonnegative, we have $$\{\begin{array}{l}0\le x\\ 0\le 18-2x& \Rightarrow \\ 0\le 48-2x\end{array}0\le x\le 9.$$ Because $V$ is continuous on the interval $[0,9]$, its absolute maximum occurs either at an endpoint $x=0$, $x=9$, or at a critical point. To obtain critical points $$\frac{dV}{dx}=12x^2-264x+864=0$$ $$\Rightarrow x^2-22x+72=(x-4)(x-18)=0$$ $$\Rightarrow x=4,\text{ or }x=18.$$ However, $x=18\notin [0,9]$. Therefore, the only critical point in the interval $[0,9]$ is $x=4$, and the absolute maximum of $V$ is one of the following numbers.

	$0$	$4$	$9$
$V(x)$	$0$	$1600$	$0$

Therefore, the maximum volume is obtained when $x=4$.

The graph of $V(x)$ is shown in Figure 6.

The distance between the point $(0,2)$ and a point $(x,y)$ is $$d=\sqrt{(x-0{)}^2+(y-2{)}^2.}$$ Because $(x,y)$ is on the graph of $x^2+8=-4y$, then $$(x,y)=(x,-\frac{1}{4}(x^2+8))$$ and the distance $d$ is $$\begin{array}{rl}d& =\sqrt{x^2+{(\frac{1}{2}(8-x^2)-2)}^2}\\ & =\sqrt{x^2+{(2-\frac{1}{2}{x}^2)}^2}.\end{array}$$ We notice that the $x$ value that minimizes $d$ also minimizes the expression inside the radical. So instead of minimizing $d$, we can minimize $d^2$ (that is, the expression inside the radical) to avoid differentiation of a square root. Therefore, the function $f$ to be minimized is $$\begin{array}{rl}f(x)& =x^2+{(2-\frac{1}{2}{x}^2)}^2\\ & =x^2+4-2x^2+\frac{1}{4}{x}^4\\ & =\frac{1}{4}{x}^4-x^2+4.\end{array}$$ Because $f(x)$ is defined for all real $x$, there are no endpoints of the domain to consider. The derivative of $f$ $$f^{\prime }(x)=x^3-2x=x(x^2-2)$$ is zero when $$x=0,\pm \sqrt{2}.$$

To apply the First Derivative Test, we need to determine the sign of $f^{\prime }$:

	$-\mathrm{\infty }$		$-\sqrt{2}$		$0$		$\sqrt{2}$		$+\mathrm{\infty }$
sign of $x+\sqrt{2}$		$-–-$	$0$	$+++$	$+$	$+++$	$+$	$+++$
sign of $x$		$-–-$	$-$	$-–-$	$0$	$+++$	$+$	$+++$
sign of $x-\sqrt{2}$		$-–-$	$-$	$-–-$	$-$	$—$	$+$	$+++$
$\therefore$ sign of $f^{\prime }(x)$		$-–-$	$0$	$+++$	$0$	$—$	$0$	$+++$
Increasing/Decreasing $f(x)$		$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\searrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2198.svg">$		$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$		$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\searrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2198.svg">$		$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$

It follows from the First Derivative Test and the above table that $x=0$ yields a local maximum, whereas $x=\pm \sqrt{2}$ yield a minimum distance. $${d|}_{x=\pm \sqrt{2}}=\sqrt{2+(2-\frac{1}{2}2{)}^2}=\sqrt{3}.$$

Let’s see what will happen if we use the Second Derivative Test instead of the First Derivative Test $$f^{\prime \prime }(x)=x^2-2$$ Because $f^{\prime \prime }(0)=-2<0$, $f$ has a local maximum at $x=0$. Because$f^{\prime \prime }(\pm \sqrt{2})=0$, the Second Derivative Test fails, and we cannot determine if $f$ has a local maximum, a local minimum at $x=\pm \sqrt{2}$.

Therefore, $f(\pm \sqrt{2})=\sqrt{3}$ is the shortest distance from the point $(0,2)$ to the parabola $8-x^2=2y$, and the closest points on the parabola to $(0,2)$ are $(\sqrt{2},3)$ and $(-\sqrt{2},3)$.

The graph of $d=\sqrt{x^2+(2-x^2/2{)}^2}$ is shown in Figure 8.

Let
$r=$ radius of the cylinder
$h=$ height of the cylinder
$V=$ volume of the cylinder
We know $$\begin{array}{}\text{(i)}& V=\pi r^{2}h,\end{array}$$ and we want to maximize $V$ provided that the cylinder is inscribed in the cone. Here $V$ depends both on $r$ and $h$ and it is necessary to express $V$ in only one variable (either $r$ or $h$). To do so, we look for a relationship that connects $r$ and $h$. If we look at the cone from the side, we realize that $\mathrm{△}OAB$ and $\mathrm{△}{O}^{\prime }A{B}^{\prime }$ are similar (Figure 10).

 

$$\stackrel{\mathrm{△}}{OAB}\sim \stackrel{\mathrm{△}}{O^{\prime }A{B}^{\prime }}$$ $$\Rightarrow \frac{A{O}^{\prime }}{AO}=\frac{O^{\prime }{B}^{\prime }}{OB}$$ or $$\frac{30-h}{30}=\frac{r}{18}$$ $$\Rightarrow 30(18)-18h=30r$$ $$\begin{array}{}\text{(ii)}& h=30-\frac{5}{3}r\end{array}$$ Substituting (ii) into (i), we obtain $$V=\pi r^2(30-\frac{5}{3}r)=30\pi r^2-\frac{5}{3}\pi r^3.$$ Now $V$ is a function of $r$ alone. Because $r$ may vary only between 0 and 18 (the radius of the cone) $$0\le r\le 18,$$ and $V$ is a continuous function, the existence of a maximum value is guaranteed. The maximum value occurs either at $r=0$, $r=18,$ or at a critical number. $$\begin{array}{rl}{V}^{\prime }(r)& =60\pi r-5\pi r^2\\ & =5\pi r(12-r).\end{array}$$ The solutions of $V^{\prime }(r)=0$ are $$r=0,r=12.$$ It suffices to compute $V(r)$ at the critical numbers and the endpoints

$r$	$0$	$12$	$18$
$V$	$0$	$1440\pi$	$0$

So the maximum volume occurs when $r=12$ and $h=30-\frac{5}{3}(12)=10$.

The graph of $V(r)$ is shown in Figure 11.

Let
$x=$ length of $BC$
$y=$ length of $CE$
$L=$ length of $BE$

 

From $\mathrm{△}BCE$ $$\begin{array}{}\text{(i)}& L^2=x^2+y^2.\end{array}$$ We want to minimize $L$, but $L$ depends on two variables $x$ and $y$. To eliminate one of the variables, we need a relationship between $x$ and $y$. From $\mathrm{△}BAD$, we have $$A{B}^2+A{D}^2=B{D}^2.$$ Because before folding the page, $\mathrm{△}BED$ was $\mathrm{△}BEC$, we have $BD=BC$ and thus $$(a-x{)}^2+A{D}^2=x^2$$ $$\Rightarrow A{D}^2=x^2-(a-x{)}^2=2ax-a^2$$ Because $AD=C{D}^{\prime }=2ax-x^2$, from $\mathrm{△}D{D}^{\prime }E$ $$D{D}^{\prime 2}+D^{\prime }{E}^2=D{E}^2$$ and $$a^2+(y-AD{)}^2=y^2$$ or $$a^2+(y-\sqrt{2ax-a^2}{)}^2=y^2$$ $$\Rightarrow -2y\sqrt{2ax-a^2}+2ax=0$$ $$\begin{array}{}\text{(ii)}& \Rightarrow y^2=\frac{a^2{x}^2}{2ax-a^2}=\frac{a{x}^2}{2x-a}\end{array}$$ Substituting (ii) in (i) we obtain $$L^2=x^2+\frac{a{x}^2}{2x-a}.$$ Instead of minimizing $L$, we can minimize $L^2$ to avoid differentiation of a radical. Let $$\begin{array}{rl}f(x)& =x^2+\frac{a{x}^2}{2x-a}\\ & =\frac{2x^3-a{x}^2+a{x}^2}{2x-a}\\ & =\frac{2x^3}{2x-a}.\end{array}$$ The critical number (or critical point) of $f$ is obtained by setting $f^{\prime }(x)=0$: $$f^{\prime }(x)=\frac{6x^2(2x-a)-4x^3}{(2x-a{)}^2}=0$$ $$\Rightarrow 8x^3-6a{x}^2=2x^2(4x-3a)=0$$

$$\Rightarrow x=0,x=\frac{3a}{4}.$$ Note that $x$ must be larger than $a/2$, otherwise the expression under the radical $L=\sqrt{2x^3/(2x-a)}$ becomes negative. Therefore, $x=0$ is not acceptable. Because $(2x-a{)}^2>0$, the sign of $f^{\prime }(x)$ is the same as the sign of its numerator, and the numerator $8x^3-6a{x}^2=2x^2(4x-3a)$ is negative if $x<3a/4$ and is positive if $x>3a/4$. Therefore, by the First Derivative Test, $x=3a/4$ is a local minimum point ($f$ varies from a decreasing function to an increasing one), and $L$ for $x=3a/4$ is $${\sqrt{\frac{2x^3}{2x-a}}|}_{x=\frac{3a}{4}}=\frac{3\sqrt{3}}{4}a\approx 1.299a.$$ If $x=a$, then $$L=\sqrt{\frac{2a^2}{2a-a}}=\sqrt{2}a\approx 1.414a$$ and $$\underset{x\to a/2^{+}}{lim}L=\underset{x\to a/2^{+}}{lim}\sqrt{\frac{2x^3}{2x-a}}\stackrel{\left[\frac{2a^3}{0^{+}}\right]}{=}+\mathrm{\infty }.$$ These calculations show that the absolute minimum of $L$ is $3\sqrt{3}a/4$.

The graph of $L/a$ versus $x/a$ is shown in Figure 14. $$\begin{array}{rl}\frac{L}{a}& =\frac{1}{a}\sqrt{\frac{2x^3}{2x-a}}\\ & =\sqrt{\frac{2x^3}{a^3(2\frac{x}{a}-1)}}\\ & =\sqrt{\frac{2(x/a{)}^3}{2(x/a)-1}}\end{array}$$

 

(b) The folded area $A$ is $$\begin{array}{}\text{(iii)}& A=xy\end{array}$$ Substituting (ii) in (iii), we obtain $$\begin{array}{rl}A& =x\sqrt{\frac{a{x}^2}{2x-a}}\\ & =\sqrt{\frac{a{x}^4}{2x-a}}.\end{array}$$ Again to avoid differentiation of a radical, we can minimize $A^2$ instead of minimizing $A$. So we minimize $$g(x)=\frac{a{x}^4}{2x-a}.$$ The critical point is obtained by setting $$g^{\prime }(x)=\frac{4a{x}^3(2x-a)-2a{x}^4}{2x-a}=0$$

$$\Rightarrow 2a{x}^3[2(2x-a)-x]=0$$ $$\Rightarrow x=0,x=\frac{2a}{3}.$$ Again $x=0$ is not within the domain of $A$, which is the interval $(a/2,a]$. When $x=2a/3$, the area is $$A=\frac{2a}{3}\sqrt{\frac{a\frac{4}{9}{a}^2}{\frac{4a}{3}-a}}=\frac{4a^2}{3\sqrt{3}}.$$ We can show that $A\to \mathrm{\infty }$ as $x\to +\mathrm{\infty }$ or $x\to a/2^{+}$, and $${A|}_{x=a}=a^2.$$ Therefore, $A$ has an absolute minimum when $x=2a/3$.

The graph of $A/a^2$ versus $x/a$ is shown in Figure 15.

From Figure 17, the length of the ladder is $$\begin{array}{rl}L& =L_1+L_2\\ & =\sqrt{x^2+b^2}+\sqrt{y^2+a^2}\end{array}$$ Here $L$ depends on both $x$ and $y$. To make $L$ a function of one variable only, we need a relationship between $x$ and $y$.

 

From $\mathrm{△}ABC$: $$\tan \theta =\frac{b}{x}$$ and from $\mathrm{△}ADE$: $$\tan \theta =\frac{y}{a}.$$ So the relationship between $x$ and $y$ is $$\frac{b}{x}=\frac{y}{a}\Rightarrow y=\frac{ab}{x}.$$ Therefore, the length of the ladder can be expressed as $$\begin{array}{rl}L& =\sqrt{x^2+b^2}+\sqrt{\frac{a^2{b}^2}{x^2}+a^2}\\ & =\sqrt{x^2+b^2}+\sqrt{\frac{a^2{b}^2+a^2{x}^2}{x^2}}\\ & =\sqrt{x^2+b^2}+\frac{a}{x}\sqrt{b^2+x^2}\\ & =\sqrt{x^2+b^2}(1+\frac{a}{x}).\end{array}$$ The critical point of $L$ is obtained by setting $dL/dx=0$: $$\begin{array}{rl}\frac{dL}{dx}& =\frac{2x}{2\sqrt{x^2+b^2}}(1+\frac{a}{x})-\frac{a}{x^2}\sqrt{x^2+b^2}\\ & =\frac{1}{\sqrt{x^2+b^2}}[x+a-\frac{a}{x^2}(x^2+b^2)]\end{array}$$ $$\frac{dL}{dx}=0\Rightarrow x-\frac{a{b}^2}{x^2}=0\Rightarrow x=\sqrt[3]{a{b}^2}.$$ and the minimum length $L$ is $$\begin{array}{rl}{\sqrt{x^2+b^2}(1+\frac{a}{x})|}_{x=\sqrt[3]{a{b}^2}}& =\sqrt{a^{2/3}{b}^{4/3}+b^2}(1+\frac{a^{2/3}}{b^{2/3}})\\ & =\sqrt{\frac{a^{2/3}{b}^{4/3}+b^2}{b^{4/3}}}(b^{2/3}+a^{2/3})\\ & =(a^{2/3}+b^{2/3}{)}^{3/2}.\end{array}$$ Obviously as $x\to 0$ or $x\to +\mathrm{\infty }$, $L\to +\mathrm{\infty }$, and the local minimum of $L$ is also its absolute minimum.