# 4.7 Maxima/minima problems  (Page 6/10)

 Page 6 / 10

Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function

$f\left(x,y\right)=4{x}^{2}-2xy+6{y}^{2}-8x+2y+3$

on the domain defined by $0\le x\le 2$ and $-1\le y\le 3.$

The absolute minimum occurs at $\left(1,0\right)\text{:}$ $f\left(1,0\right)=-1.$

The absolute maximum occurs at $\left(0,3\right)\text{:}$ $f\left(0,3\right)=63.$

## Chapter opener: profitable golf balls

Pro- $\text{T}$ company has developed a profit model that depends on the number x of golf balls sold per month (measured in thousands), and the number of hours per month of advertising y , according to the function

$z=f\left(x,y\right)=48x+96y-{x}^{2}-2xy-9{y}^{2},$

where $z$ is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is $50,000,$ and the maximum number of hours of advertising that can be purchased is $25.$ Find the values of $x$ and $y$ that maximize profit, and find the maximum profit.

Using the problem-solving strategy, step $1$ involves finding the critical points of $f$ on its domain. Therefore, we first calculate ${f}_{x}\left(x,y\right)$ and ${f}_{y}\left(x,y\right),$ then set them each equal to zero:

$\begin{array}{ccc}\hfill {f}_{x}\left(x,y\right)& =\hfill & 48-2x-2y\hfill \\ \hfill {f}_{y}\left(x,y\right)& =\hfill & 96-2x-18y.\hfill \end{array}$

Setting them equal to zero yields the system of equations

$\begin{array}{ccc}\hfill 48-2x-2y& =\hfill & 0\hfill \\ \hfill 96-2x-18y& =\hfill & 0.\hfill \end{array}$

The solution to this system is $x=21$ and $y=3.$ Therefore $\left(21,3\right)$ is a critical point of $f.$ Calculating $f\left(21,3\right)$ gives $f\left(21,3\right)=48\left(21\right)+96\left(3\right)-{21}^{2}-2\left(21\right)\left(3\right)-9{\left(3\right)}^{2}=648.$

The domain of this function is $0\le x\le 50$ and $0\le y\le 25$ as shown in the following graph.

${L}_{1}$ is the line segment connecting $\left(0,0\right)$ and $\left(50,0\right),$ and it can be parameterized by the equations $x\left(t\right)=t,y\left(t\right)=0$ for $0\le t\le 50.$ We then define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)\text{:}$

$\begin{array}{cc}\hfill g\left(t\right)& =f\left(x\left(t\right),y\left(t\right)\right)\hfill \\ & =f\left(t,0\right)\hfill \\ & =48t+96\left(0\right)-{y}^{2}-2\left(t\right)\left(0\right)-9{\left(0\right)}^{2}\hfill \\ & =48t-{t}^{2}.\hfill \end{array}$

Setting ${g}^{\prime }\left(t\right)=0$ yields the critical point $t=24,$ which corresponds to the point $\left(24,0\right)$ in the domain of $f.$ Calculating $f\left(24,0\right)$ gives $576.$

${L}_{2}$ is the line segment connecting  and $\left(50,25\right),$ and it can be parameterized by the equations $x\left(t\right)=50,y\left(t\right)=t$ for $0\le t\le 25.$ Once again, we define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)\text{:}$

$\begin{array}{cc}\hfill g\left(t\right)& =f\left(x\left(t\right),y\left(t\right)\right)\hfill \\ & =f\left(50,t\right)\hfill \\ & =48\left(50\right)+96t-{50}^{2}-2\left(50\right)t-9{t}^{2}\hfill \\ & =-9{t}^{2}-4t-100.\hfill \end{array}$

This function has a critical point at $t=-\frac{2}{9},$ which corresponds to the point $\left(50,-\frac{2}{9}\right).$ This point is not in the domain of $f.$

${L}_{3}$ is the line segment connecting $\left(0,25\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(50,25\right),$ and it can be parameterized by the equations $x\left(t\right)=t,y\left(t\right)=25$ for $0\le t\le 50.$ We define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)\text{:}$

$\begin{array}{cc}\hfill g\left(t\right)& =f\left(x\left(t\right),y\left(t\right)\right)\hfill \\ & =f\left(t,25\right)\hfill \\ & =48t+96\left(25\right)-{t}^{2}-2t\left(25\right)-9\left({25}^{2}\right)\hfill \\ & =\text{−}{t}^{2}-2t-3225.\hfill \end{array}$

This function has a critical point at $t=-1,$ which corresponds to the point $\left(-1,25\right),$ which is not in the domain.

${L}_{4}$ is the line segment connecting $\left(0,0\right)\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}\left(0,25\right),$ and it can be parameterized by the equations $x\left(t\right)=0,y\left(t\right)=t$ for $0\le t\le 25.$ We define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)\text{:}$

$\begin{array}{cc}\hfill g\left(t\right)& =f\left(x\left(t\right),y\left(t\right)\right)\hfill \\ & =f\left(0,t\right)\hfill \\ & =48\left(0\right)+96t-{\left(0\right)}^{2}-2\left(0\right)t-9{t}^{2}\hfill \\ & =96t-{t}^{2}.\hfill \end{array}$

This function has a critical point at $t=\frac{16}{3},$ which corresponds to the point $\left(0,\frac{16}{3}\right),$ which is on the boundary of the domain. Calculating $f\left(0,\frac{16}{3}\right)$ gives $256.$

We also need to find the values of $f\left(x,y\right)$ at the corners of its domain. These corners are located at $\left(0,0\right),\left(50,0\right),\left(50,25\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(0,25\right)\text{:}$

$\begin{array}{ccc}\hfill f\left(0,0\right)& =\hfill & 48\left(0\right)+96\left(0\right)-{\left(0\right)}^{2}-2\left(0\right)\left(0\right)-9{\left(0\right)}^{2}=0\hfill \\ \hfill f\left(50,0\right)& =\hfill & 48\left(50\right)+96\left(0\right)-{\left(50\right)}^{2}-2\left(50\right)\left(0\right)-9{\left(0\right)}^{2}=-100\hfill \\ \hfill f\left(50,25\right)& =\hfill & 48\left(50\right)+96\left(25\right)-{\left(50\right)}^{2}-2\left(50\right)\left(25\right)-9{\left(25\right)}^{2}=-5825\hfill \\ \hfill f\left(0,25\right)& =\hfill & 48\left(0\right)+96\left(25\right)-{\left(0\right)}^{2}-2\left(0\right)\left(25\right)-9{\left(25\right)}^{2}=-3225.\hfill \end{array}$

The maximum critical value is $648,$ which occurs at $\left(21,3\right).$ Therefore, a maximum profit of $\text{}648,000$ is realized when $21,000$ golf balls are sold and $3$ hours of advertising are purchased per month as shown in the following figure.

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