Optimization

Section 6.8 Optimization

These answers correspond to the problems in Section 3.3.

Note that the function $f$ is continuous on the closed interval $[-1,2]\text{.}$ By the Intermediate Value Theorem the function $f$ attains its maximum and minimum values on $[-1,2]\text{.}$ To find those global extrema we evaluate and compare the values of $f$ at the endpoints and critical numbers that belong to $(-1,2)\text{.}$ From $f'(x)=6x-9=3(2x-3)$ we conclude that the critical number is $\ds x=\frac{3}{2}\text{.}$ From $f(-1)=12\text{,}$ $f(2)=-6\text{,}$ and $\ds f\left( \frac{3}{2}\right) =-\frac{27}{4}$ we conclude that the maximum value is $f(-1)=12$ and the minimum value is $\ds f\left( \frac{3}{2}\right) =-\frac{27}{4}\text{.}$
The global minimum value is $f(-4)=f(2)=-21\text{,}$ and the global maximum value is $f(6)=139\text{.}$ Note that $f(2)=-21$ is also a local minimum and that $f(-2)$ is a local maximum. (Reminder: By our definition, for $x=c$ to be a local extremum of a function $f$ it is necessary that $c$ is an interior point of the domain of $f\text{.}$ This means that there is an open interval $I$ contained in the domain of $f$ such that $c\in I\text{.}$)
From $f^\prime(x)=ax^{a-1}(1-x)^b-bx^a(1-x)^{b-1}=x^{a-1}(1-x)^{b-1}(a-(a+b)x)$ and the fact that $a$ and $b$ are positive conclude that $\ds x=\frac{a}{a+b}\in (0,1)$ is a critical number of the function $f\text{.}$ Since $f(0)=f(1)=0$ and $f(x)>0$ for all $x\in (0,1)$ it follows that the maximum value of $f$ is $\ds f\left( \frac{a}{a+b}\right) =\left( \frac{a}{a+b}\right)^a\left( \frac{b}{a+b}\right)^b\text{.}$
From $\ds f(x)=\left\{ \begin{array}{rr} 3x-5\amp \mbox{if } x\geq \frac{5}{3}\\ -3x+5\amp \mbox{if } x\lt \frac{5}{3} \end{array} \right.$ we conclude that $\ds f'(x)=\left\{ \begin{array}{rr} 3\amp \mbox{if } x> \frac{5}{3}\\ -3\amp \mbox{if } x\lt \frac{5}{3} \end{array} \right.\text{.}$ Thus, for $\ds x\not= \frac{5}{3}\text{,}$ $f'(x)\not= 0$ and the derivative of $f$ is not defined at $\ds x= \frac{5}{3}\text{.}$ We conclude that the only critical number of the function $f$ on the interval $[-3,2]$ is $\ds x= \frac{5}{3}\text{.}$ Clearly, $\ds f\left( \frac{5}{3}\right) =0\text{.}$ From $f(-3)=14$ and $f(1)=2$ it follows that the global and local minimum is $\ds f\left( \frac{5}{3}\right) =0$ and that the global maximum is $f(-3)=14\text{.}$
The question is to find the minimum value of the function $f(x)=x^2+(12-x)^2\text{,}$ $x\in (0,12)\text{.}$ From $f'(x)=4(x-6)$ it follows that $x=6$ is the only critical number. From $f"(6)=4>0\text{,}$ by the second derivative test, it follows that $f(6)=72$ is the minimum value of the function $f\text{.}$
Note that $f(0)=f(1)=0$ and that $f(x)>0$ for $x\in (0,1)\text{.}$ Thus by the Intermediate Value Theorem there is $c\in (0,1)$ such that $f(c)$ is the maximum value of $f\text{.}$ Since $f$ is differentiable on $(0,1)\text{,}$ $c$ must be a critical point. Note that $f'(x)=x^{a-1}(1-x)^{b-1}(a-(a+b)x)\text{.}$ Since $a$ and $b$ are both positive we have that $\ds x=\frac{a}{a+b}\in (0,1)\text{.}$ Thus $\ds x=\frac{a}{a+b}$ is the only critical point of the function $f$ in the interval $(0,1)$ and $\ds f\left( \frac{a}{a+b}\right) =\frac{a^ab^b}{(a+b)^{a+b}}$ is the maximum value.
The distance between a point $(x,y)$ on the curve and the point $(0,-3)$ is $\ds d=\sqrt{(x-0)^2+(y-(-3))^2}=\sqrt{y^4+(y+3)^2}\text{.}$ The question is to minimize the function $f(y)= y^4+(y+3)^2\text{,}$ $y\in \mathbb{R}\text{.}$ From $f'(y)=2(2y^3+y+3)=2(y+1)(2y^2-2y+3)$we conclude that $y=-1$ is the only critical number of the function $f\text{.}$ From $f"(-1)=10>0\text{,}$ by the second derivative test we conclude that $f(-1)=5$ is the (local and global) minimum value of $f\text{.}$ Thus the closest point is $(-1,-1)\text{.}$
Let $x$ be the radius of the circle. The question is to minimize the function $\ds f(x)=\pi x^2+\left( \frac{40-2\pi x}{4}\right) ^2\text{,}$ $\ds x\in \left( 0,\frac{20}{\pi }\right)\text{.}$ (We are given that there are TWO pieces.) The only critical number of the function $f$ is $\ds x=\frac{20}{\pi +4}\text{.}$ To minimize the total area the two pieces should be of the length $\ds \frac{40\pi }{\pi +4}$ and $\ds \frac{160 }{\pi +4}\text{.}$
$\ds x=3$ and $y=2\text{.}$ The question is to minimize the function $\ds f(x)=x^2+\frac{3(7-x)^2}{4}\text{,}$ $x\in (0,7)\text{.}$ (We are given that there are TWO pieces.) The critical point is $x=3\text{.}$
The question is to maximize the function $\ds f(x)=\frac{x^2\sqrt{3}}{36}+\frac{(4-x)^2}{16}\text{,}$ $x\in [0,4]\text{.}$ Note that $f"(x)>0$ for $x\in (0,4)$ and conclude that the maximum value must occur at $x=0$ and/or $x=4\text{.}$ Since $f(4)\lt f(0)\text{,}$ the maximum total area is obtained whenonly the square is constructed.
Maximize the function $P(x)=2x+\sqrt{R^2-x^2}\text{,}$ where $2x$ represents the width of a rectangle inscribed in the semicircle.
1. See Figure 6.8.1.
  
  Figure 6.8.1. Rectangle inscribed in the region $y=4$ and $y=x^2$
2. $P=4a+2(4-a^2)=2(4+2a-a^2)\text{,}$ $a\in (0,1)\text{.}$
3. From $\ds \frac{dP}{da}=4(1-a)$ it follows that $\ds a=1$ is the only critical number. The fact that $\ds f"(a) =-8\lt 0$ for all $a\in (0,1)$ implies, by the second derivative test, that $\ds P(1)$ is the maximum value.
4. $\ds P(1)=10\text{.}$
Let $(x,0)$ be the bottom right vertex of the rectangle. The question is to maximize $f(x)=2x(12-x^2)\text{,}$ $x\in (0,2\sqrt{3})\text{.}$ The only critical number is $x=2\text{.}$ The length of the rectangle with the largest area is 4 and its height is 8.
Let $x$ be the length of one side of the fence that is perpendicular to the wall. Note that the length of the side of the fence that is parallel to the wall equals $400-2x$ and that this number cannot be larger than 100.The question is to maximize the function $f(x)=x(400-2x)\text{,}$ $x\in [150,400)$ . The only solution of the equation $f'(x)=4(100-x)=0$ is $x=100$ but this value is not in the domain of the function $f\text{.}$ Clearly $f'(x)\lt 0$ for $x\in [150,400)$ which implies that $f$ is decreasing on its domain. Therefore the maximum area that can be enclosed is $f(150)=15000$ ft$^2\text{.}$
$L=15\sqrt{3}\text{.}$ To minimize $L^2=(x+5)^2+y^2\text{,}$ use the fact that $\ds \frac{x}{10\sqrt{2}}=\frac{x+5}{y}$ and the first derivative.
(a) $\ds \alpha =\tan^{-1}\left(\frac{7}{x}\right)- \tan^{-1}\left(\frac{5}{x}\right)\text{.}$ (b) $x=\sqrt{35}\text{.}$
Let $\ds (x,y)=\left( x,\frac{b}{a}\sqrt{a^2-x^2}\right)$ be the upper right vertex of the rectangle. The question is to maximize the function $\ds f(x)=\frac{4b}{a}x\sqrt{a^2-x^2}\text{,}$ $x\in (0,a)\text{.}$ From $\ds f'(x)=\frac{4b}{a}\frac{a^2-2x^2}{\sqrt{a^2-x^2}}$ we conclude that the only critical number is $\ds x=\frac{a}{\sqrt{2}}\text{.}$ By the first derivative test, there is a local maximum at this critical number. Since $\ds \lim _{x\to 0^+}f(x)= \lim _{x\to a^-}f(x)=0\text{,}$ it follows that $\ds f\left( \frac{a}{\sqrt{2}}\right) =2ab$ is the maximum value of the function $f\text{.}$ Thus to maximize the area of the soccer field its length should be $a\sqrt{2}$ and its width should be $b\sqrt{2}\text{.}$
Let $a$ be the length of the printed material on the poster. Then the width of this area equals $\ds b=\frac{384}{a}\text{.}$ It follows that the length of the poster is $x=a+8$ and the width of the poster is $\ds y=b+12=\frac{384}{a}+12\text{.}$ The question is to minimize the function $\ds f(a)=xy=(a+8)\left( \frac{384}{a}+12\right) =12\left( 40+a+\frac{256}{a}\right)\text{.}$ It follows that the function has a local minimum at $\ds a=16\text{.}$ The dimensions of the poster with the smallest area are $x=24$ cm and $y=36$ cm.
$2\sqrt{30}\cdot 3\sqrt{30}\text{.}$ Maximize the function $\ds A(x)=(x-2)\left( \frac{180}{x}-3\right)\text{,}$ where $x$ represents the length of the poster.
$(\sqrt{15}+2)\times (2\sqrt{15}+4)\text{.}$
Let $P$ be the point on the shore where Maya lands her boat and let $x$ be the distance from $P$ to the point on the shore that is closest to her initial position. Thus to reach the village she needs to row the distance $z=\sqrt{4+x^2}$ and run the distance $y=6-x\text{.}$ Time needed to row the distance $z$ is given by $\ds T_1=\frac{z}{2}$ and time she needs to run is $\ds T_2=\frac{y}{5}\text{.}$ Therefore the question is to minimize the function $\ds T=T(x)=T_1+T_2=\frac{\sqrt{4+x^2}}{2}+\frac{6-x}{5}\text{,}$ $x\in [0,6]\text{.}$ From $\ds f'(x)=\frac{x}{2\sqrt{4+x^2}}-\frac{1}{5}$ it follows that the only critical number is $\ds x=\frac{4}{3}\text{.}$ From $\ds T(0)=\frac{11}{5}=2.2\text{,}$ $T(6)=\sqrt{10}\text{,}$ and $\ds T\left( \frac{4}{3}\right) \approx 2.135183758$ it follows that the minimum value is $\ds T\left( \frac{4}{3}\right)\text{.}$ Maya should land her boat $\ds \frac{4}{3}$ km from the point initially nearest to the boat.
$x=10-\sqrt{3}\text{.}$ Minimize the function $T(x)=\frac{1}{2}\cdot \sqrt{9+(10-x)^2}+\frac{x}{4}\text{,}$ where $x$ is the distance between a point on the road and the store.
1. Let $y$ be the height of the box. Then the surface area is given by $S=2x^2+4xy\text{.}$ From $S=150$ it follows that $\ds y=\frac{1}{2}\left( \frac{75}{x}-x\right)\text{.}$ Therefore the volume of the box is given by $\ds V=V(x)=\frac{x}{2}\left( 75-x^2\right)\text{.}$
2. From the fact that $\ds y=\frac{1}{2}\left( \frac{75}{x}-x\right) >0$ it follows that the domain of the function $V=V(x)$ is the interval $[1, 5\sqrt{3})\text{.}$
3. Note that $\ds \frac{dV}{dx}=\frac{3}{2}(25-x^2)$ and that $\ds \frac{d^2V}{dx^2}=-3x\lt 0$ for all $x\in (1, 5\sqrt{3})\text{.}$ Thus the maximum value is $V(5)=125$ cube units.
1. Note that $\ds y=\frac{10}{x^2}\text{.}$ The cost function is given by $\ds C(x)=5x^2+2\cdot 4\cdot x\cdot \frac{10}{x^2}=5x^2+\frac{80}{x}\text{,}$ $x>0\text{.}$
2. $\ds 2\times 2\times \frac{5}{2}\text{.}$ The minimum cost is $C(2)=\$ 60\text{.}$
Let $x$ be the length and the width of the box. Then its height is given by $\ds y=\frac{13500}{x^2}\text{.}$ It follows that the surface area is $\ds S=x^2+\frac{54000}{x}$ cm$^2\text{,}$ $x>0\text{.}$ The question is to minimize $S\text{.}$ From $\ds \frac{dS}{dx}=2x-\frac{54000}{x^2}$ and $\ds \frac{d^2S}{dx^2}=2+\frac{3\cdot 54000}{x^3}>0$ for all $x>0$ it follows that the function $S$ has a local and global minimum at $x=30\text{.}$
We need to maximize the area of the trapezoid with parallel sides of lengths $a=2$ and $c=2+2\cdot 2\cos \theta=2+4\cos\theta$ and the height $h=2\sin \theta\text{.}$ Thus we maximize the function $\ds A=A(\theta )=\frac{2+(2+4\cos \theta )}{2}\cdot 2\sin \theta=4(\sin \theta +\sin \theta \cos\theta)\text{,}$ $\theta \in (0,\pi )\text{.}$ From $\ds \frac{dA}{d\theta }=4(\cos \theta +\cos ^2\theta-\sin ^2\theta)=4(2\cos ^2\theta +\cos \theta -1)=4(2\cos \theta -1)(\cos \theta +1)$ we obtain the critical number $\ds \theta =\frac{\pi }{3}\text{.}$ The First Derivative Test confirms that $\ds \theta =\frac{\pi }{3}$ maximizes the cross sectional area of the trough.
$\ds r=\frac{16}{3}$ cm, $h=4$ cm.
Let $r$ be the radius of the base of a cylinder inscribed in the cone and let $h$ be its height. From $\ds \frac{H}{R}=\frac{h}{R-r}$ (see Figure 6.8.2) we conclude that $\ds h=\frac{H(R-r)}{R}\text{.}$

Figure 6.8.2. A cylinder and a cone
Thus the volume of the cylinder is $\ds V=V(r)=\frac{\pi H}{R}r^2(R-r)\text{,}$ $r\in (0,R)\text{.}$ From $\ds \frac{dV}{dr}=\frac{\pi H}{R}r(2R-3r)$ and $\ds \frac{d^2V}{dr^2}=\frac{2\pi H}{R}(R-3r)$ it follows that the maximum value of the volume of the cylinder is $\ds V\left( \frac{2R}{3}\right)=\frac{4\pi HR^2}{27}\text{.}$ The dimensions are $\ds r=\frac{2R}{3}$ and $\ds h=\frac{H}{3}\text{.}$
$\ds r=\sqrt{\frac{10}{3\pi }}$ m, $\ds h=\left( \frac{5}{\pi}\sqrt{\frac{3\pi}{10 }}-\frac{1}{2}\sqrt{\frac{10}{3\pi }}\right)$ m, $\ds V=\frac{10}{3}\sqrt{\frac{10}{3\pi }}$ m$^3\text{.}$
Let $r$ be the radius of the base of the pot. Then the height of the pot is $\ds h=\frac{250}{\pi r^2}\text{.}$ The cost function is $\ds C(r)=4\pi r^2+\frac{1000}{r}\text{,}$ $r>0\text{.}$ The cost function has its minimum at $\ds r=\frac{5}{\sqrt[3]{\pi }}\text{.}$
$\ds R=\frac{10}{\sqrt[3]{\pi}}$ cm, $\ds h=\frac{10}{\sqrt[3]{\pi}}$ cm.
1. The surface area of the can is $S=2\pi rh+2\pi r^2\text{.}$ The amount of material wasted is $A-S=2(4-\pi )r^2\text{.}$
2. From $V=\pi r^2h$ it follows that the amount of material needed to make a can of the given volume $V$ is $\ds A=A(r)=\frac{2V}{r}+8r^2\text{.}$ This function has its minimum at $\ds r=\frac{\sqrt[3]{V}}{2}\text{.}$ The ratio of the height to diameter for the most economical can is $\ds \frac{h}{r}=\frac{4}{\pi }\text{.}$
3. $\ds A"(r)=\frac{4V}{r^3}+8>0$ for $r>0\text{.}$
$\ds r=\sqrt[3]{20}\text{,}$ $\ds h=\frac{14}{\sqrt[3]{50}}\text{.}$ Minimize the cost function $\ds C=C(r)=7r^2\pi +\frac{280\pi }{r}\text{.}$
From Figure 6.8.3 conclude that $\ds r^2=R^2-(h-R)^2=h(4R-h)\text{.}$ Then the volume of the cone as a function of $h$ is given by $\ds V=\frac{\pi }{3}h^2(4R-h)\text{.}$ Maximize.

Figure 6.8.3. Cone inscribed in a sphere
Let $P$ be the source power of the first party's stereo and let $x$ be the distance between the person and the first party. Since the power of the second party's stereo is $64P\text{,}$ the sound level is $L(x)=kPx^{-2}+64kP(100-x)^{-2}\text{,}$ $x\in (0,100)\text{.}$ From $ds \frac{dL}{dx}=2kP\left( \frac{64}{(100-x)^3}-\frac{1}{x^3}\right)$ it follows that $x=20$ is the only critical number for the function $L\text{.}$ Since for $x\in (0,100)$

\begin{equation*} L'(x)>0\Leftrightarrow \frac{64}{(100-x)^3}-\frac{1}{x^3}>0\Leftrightarrow 64x^3>(100-x)^3 \Leftrightarrow 4x>100-x \Leftrightarrow x>20 \end{equation*}

the function $L$ is strictly increasing on the interval $(20,100)$ and strictly decreasing on the interval $(0,20)\text{.}$ Therefore, $L(20)$ is the absolute minimum.