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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]

Sagot :

GinnyAnswer
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.
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