The feasible region has four vertices: [tex]\(\{(0,0),(0,40),(40,30),(100,0)\}\)[/tex]. According to the theorem, the optimization will only occur at one of these vertices.

Now, write the optimization equation:
[tex]\[ a \cdot b + c \cdot n = \text{maximum revenue} \][/tex]

Substitute each ordered pair to determine which makes the most money:
[tex]\[
\begin{array}{l}
(0,0) \rightarrow \$ 0.00 \\
(0,40) \rightarrow 5(0) + 7.50(40) = \$ 300.00 \\
(40,30) \rightarrow 5(40) + 7.50(30) = \$ \square \\
(100,0) \rightarrow 5(100) + 7.50(0) = \$ \square \\
\end{array}
\][/tex]



Answer :

Certainly! Let's walk through the steps to determine the maximum revenue given the vertices of the feasible region and the optimization equation.

We are given the vertices of the feasible region:
[tex]\[ (0, 0), (0, 40), (40, 30), (100, 0) \][/tex]

The optimization equation is given as:
[tex]\[ 5b + 7.5n \][/tex]

Now, we will substitute each ordered pair into this equation to determine the revenue generated at each vertex:

1. Vertex (0, 0):
[tex]\[ 5(0) + 7.5(0) = 0 \][/tex]
Revenue: \[tex]$0.00 2. Vertex (0, 40): \[ 5(0) + 7.5(40) = 0 + 300 = 300 \] Revenue: \$[/tex]300.00

3. Vertex (40, 30):
[tex]\[ 5(40) + 7.5(30) = 200 + 225 = 425 \][/tex]
Revenue: \[tex]$425.00 4. Vertex (100, 0): \[ 5(100) + 7.5(0) = 500 + 0 = 500 \] Revenue: \$[/tex]500.00

To summarize:
- At vertex [tex]\((0, 0)\)[/tex], the revenue is \[tex]$0.00 - At vertex \((0, 40)\), the revenue is \$[/tex]300.00
- At vertex [tex]\((40, 30)\)[/tex], the revenue is \[tex]$425.00 - At vertex \((100, 0)\), the revenue is \$[/tex]500.00

The maximum revenue occurs at vertex [tex]\((100, 0)\)[/tex] with a revenue of \$500.00.