## Precalculus (6th Edition) Blitzer

a. $z=500x+350y$ b. $\begin{cases} x+y\leq200 \\x\geq10 \\ y\geq80 \end{cases}$ c. See graph d. $33,000$; $71,500$; $88,000$ dollars. e. $120$; $80$; $88,000$
a. Based on the given conditions, the objective function that models total daily profit can be written as $z=500x+350y$ b. Using the given constraints, we have $\begin{cases} x+y\leq200 \\x\geq10 \\ y\geq80 \end{cases}$ c. See graph and the solution region with vertices $(10,80),(10,190),(120,80)$. d. With the identified vertices, we have $z_1=500(10)+350(80)=33,000$, $z_2=500(10)+350(190)=71,500$, $z_3=500(120)+350(80)=88,000$ dollars. e. We can complete the statement as: The company will make the greatest profit by producing $120$ units of writing paper and $80$ units of newsprint each day. The maximum daily profit is $88,000$ dollars.