Applied Maximum and Minimum Problems

Practice 1: $\mathrm{V}(\mathrm{x})=\mathrm{x}(15-2 \mathrm{x})(7-2 \mathrm{x})=4 \mathrm{x}^{3}-44 \mathrm{x}^{2}+105 \mathrm{x}$
$\mathrm{V}^{\prime}(\mathrm{x})=12 \mathrm{x}^{2}-88 \mathrm{x}+105=(2 \mathrm{x}-3)(6 \mathrm{x}-35)$ which is defined for all $\mathrm{x}$ so the only critical numbers are the endpoints $x=0$ and $x=7 / 2$ and the places where $V'$ equals $0$, at $x=3 / 2$ and $x=35 / 6$ (but $35 / 6$ is not in the interval $[0,7 / 2]$ so it is not practical for this applied problem).

The maximum volume must occur when $\mathrm{x}=0, \mathrm{x}=3 / 2$, or $\mathrm{x}=7 / 2$):

$\mathrm{V}(0)=0 \cdot(15-2 \cdot 0) \cdot (7-2 \cdot 0)=0$

$\mathrm{V}\left(\frac{3}{2}\right)=\frac{3}{2} \cdot\left(15-2 \cdot \frac{3}{2}\right) \cdot\left(7-2 \cdot \frac{3}{2}\right)$

$=\frac{3}{2}(12)(4)=72 \max$

$\mathrm{V}\left(\frac{7}{2}\right)=\frac{7}{2} \cdot\left(15-2 \cdot \frac{7}{2}\right) \cdot\left(7-2 \cdot \frac{7}{2}\right)$

$=\frac{7}{2}(8)(0)=0$

Fig. 31 shows the graph of $\mathrm{V}(\mathrm{x})$.

Fig. 31

Practice 2: (a) We have $80$ feet of fencing. (See Fig. 32). Our assignment is to maximize the area of the garden: $A=x \cdot y$ (two variables). Fortunately we have the constraint that $x+y=80$ so $\mathrm{y}=80-\mathrm{x}$, and our assignment reduces to maximizing a function of one variable:

maximize $A=x \cdot y=x \cdot(80-x)=80 x-x^{2}$.

$A^{\prime}=80-2 x$ so $A^{\prime}=0$ when $x=40$.

$\mathrm{A}^{\prime \prime}=-2$ so $\mathrm{A}$ is concave down, and $\mathrm{A}$ has a maximum at $x=40$,

The maximum area is $\mathrm{A}=40 \cdot 40=1600$ square feet when $\mathrm{x}=40 \mathrm{ft}$. and $\mathrm{y}=40 \mathrm{ft}$. The maximum area garden is a square.

Fig. 32

(b) This is very similar to part (a) except we have $\mathrm{F}$ feet of fencing instead of $80$ feet. $\mathrm{x}+\mathrm{y}=\mathrm{F}$ so $\mathrm{y}=\mathrm{F}-\mathrm{x}$, and we want to maximize $\mathrm{A}=\mathrm{xy}=\mathrm{x}(\mathrm{F}-\mathrm{x})=\mathrm{Fx}-\mathrm{x}^{2}$ $\mathrm{A}^{\prime}=\mathrm{F}-2 \mathrm{x}$ so $\mathrm{A}^{\prime}=0$ when $\mathrm{x}=\mathrm{F} / 2$ and $\mathrm{y}=\mathrm{F} / 2$. The maximum area is $\mathrm{A}=\mathrm{F}^{2} / 4$ square feet and that occurs when the garden is a square and half of the new fence is used on each of the two new sides.

Practice 3: Cost $\mathrm{C}=5$ (area of top) $+3$ (area of sides) $+5$ (area of bottom) $=5\left(\pi \mathrm{r}^{2}\right)+3(2 \pi \mathrm{rh})+5\left(\pi \mathrm{r}^{2}\right)$
so our assignment is to minimize $\mathrm{C}=10 \pi \mathrm{r}^{2}+6 \pi \mathrm{rh}$, a function of two variables $\mathrm{r}$ and $\mathrm{h}$.

Fortunately we also have the constraint that volume $=1000 \mathrm{in}^{3}=\pi \mathrm{r}^{2} \mathrm{~h}$ so $\mathrm{h}=\frac{1000}{\pi \mathrm{r}^{2}}$. Then $\mathrm{C}=10 \pi \mathrm{r}^{2}+6 \pi \mathrm{r}\left(\frac{1000}{\pi \mathrm{r}^{2}}\right)=10 \pi \mathrm{r}^{2}+\frac{6000}{\mathrm{r}}$ so $\mathrm{C}^{\prime}=20 \pi \mathrm{r}-\frac{6000}{\mathrm{r}^{2}}. \mathrm{C}^{\prime}=0$ if $20 \pi \mathrm{r}-\frac{6000}{\mathrm{r}^{2}}=0$ so $20 \pi \mathrm{r}^{3}=6000$ and $\mathrm{r}=\left(\frac{6000}{20 \pi}\right)^{1 / 3} \approx 4.57$ in. Then $\mathrm{h}=\frac{1000}{\pi \mathrm{r}^{2}} \approx \frac{1000}{\pi(4.57)^{2}} \approx 15.24 \mathrm{in}$. ($\mathrm{C}^{\prime \prime}=20 \pi+\frac{12000}{\mathrm{r}^{3}} > 0$. for all $\mathrm{r} > 0$ so $\mathrm{C}$ is concave up and we have found a minimum of $C$.)