Answer
a. $z(x,y)=125x+200y$
b. $\begin{cases} x\leq450 \\ y \leq200\\ 600x+900y \leq 360,000\end{cases}$
c. See graph
d. $z(0,200)= 40,000$, $z(300,200)= 77,500$, $z(450,100)= 76,250$, $z(450,0)= 56,250$
e. $300$ rear-projection and $200$ plasma television each month; maximum monthly profit $77,500$ dollars.
Work Step by Step
a. Based on the given conditions, the total monthly profit can be written as $z(x,y)=125x+200y$
b. We can convert the constraints into inequalities as
$\begin{cases} x\leq450 \\ y \leq200\\ 600x+900y \leq 360,000\end{cases}$
c. See graph for the vertices and the solution region as a pentagon shaped area.
d. With the objective equation and vertices, we have
$z(0,0)=125(0)+200(0)=0$, $z(0,200)=125(0)+200(200)=40,000$, $z(300,200)=125(300)+200(200)=77,500$, $z(450,100)=125(450)+200(100)=76,250$, $z(450,0)=125(450)+200(0)=56,250$,
e. Based on the above results, we can complete the missing portions as: $300$ rear-projection television each month and $200$ plasma television each month. The maximum monthly profit is $77,500$ dollars.
