#### Answer

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. $120$; $80$; $88,000$

#### Work Step by Step

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.