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Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function
on the domain defined by $0\le x\le 2$ and $\mathrm{-1}\le y\le 3.$
The absolute minimum occurs at $\left(1,0\right)\text{:}$ $f\left(1,0\right)=\mathrm{-1}.$
The absolute maximum occurs at $\left(0,3\right)\text{:}$ $f\left(0,3\right)=63.$
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
where $z$ is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is $\mathrm{50,000},$ 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:
Setting them equal to zero yields the system of equations
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 $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)\text{:}$
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 $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)\text{:}$
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 $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)\text{:}$
This function has a critical point at $t=\mathrm{-1},$ which corresponds to the point $\left(\mathrm{-1},25\right),$ 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 $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right)\text{:}$
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 $\left(0,0\right),\left(50,0\right),\left(50,25\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(0,25\right)\text{:}$
The maximum critical value is $648,$ which occurs at $\left(21,3\right).$ Therefore, a maximum profit of $\text{\$}\mathrm{648,000}$ is realized when $\mathrm{21,000}$ golf balls are sold and $3$ hours of advertising are purchased per month as shown in the following figure.
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