Chapter 3.6, Problem 43E

To determine

To find: the number of newspaper and notebooks paper produced by each plant in order to maximize the profit.

Answer to Problem 43E

120 notebooks and 80 newspaper prints

Explanation of Solution

Given information:

Let $x$ be the number of notebooks and $y$ be the number of newspaper prints.

The inequalities are as follows:

  $\begin{array}{c}x+y\le 200\\ x\ge 10\\ y\ge 80\end{array}$

Let $P\left(x,y\right)$ be the maximum profit.

  $P\left(x,y\right)=400x+350y$ .

Calculation:

Use a graphing calculator to plot the inequalities as shown below.

  

The vertices are $\left(10,80\right)\text{,}\left(10,190\right)\text{ and }\left(120,80\right)$ . Now test each vertex as shown below.

  $\begin{array}{c}P\left(10,80\right)=400\left(10\right)+350\left(80\right)\\ =\$32000\\ P\left(10,190\right)=400\left(10\right)+350\left(190\right)\\ =\$70500\\ P\left(120,80\right)=400\left(120\right)+350\left(80\right)\\ =\$76000\end{array}$

Thus, the number of notebooks and newsprint production in order to maximize the profit is 120 and 80 respectively.