Linear Programming

James is trying to expand his pastry business to include cupcakes and personal cakes. He has 40 hours available to decorate the new items and can use no more than 22 pounds of cake mix. Each personal cake requires 2 pounds of cake mix and 2 hours to decorate. Each cupcake order requires one pound of cake mix and 4 hours to decorate. If he can sell each personal cake for $14.99 and each cupcake order for $16.99, how many personal cakes and cupcake orders should James make to make the most revenue?

There are four inequalities in this situation. First, state the variables. Let \(\begin{align*}p=\end{align*}\) the number of personal cakes and \(\begin{align*}c=\end{align*}\) the number of cupcake orders.

Translate this into a system of inequalities.

\(\begin{align*}2p+1c \le 22\end{align*}\) – This is the amount of available cake mix.

\(\begin{align*}2p+4c \le 40\end{align*}\) – This is the available time to decorate.

\(\begin{align*}p \ge 0\end{align*}\) – You cannot make negative personal cakes.

\(\begin{align*}c \ge 0\end{align*}\) – You cannot make negative cupcake orders.

Now graph each inequality and determine the feasible region.

The feasible region has four vertices: {(0, 0),(0, 10),(11, 0),(8, 6)}. According to our theorem, the optimization answer will only occur at one of these vertices.

Write the optimization equation. How much of each type of order should James make to bring in the most revenue?

\(\begin{align*}14.99p+16.99c=maximum \ revenue\end{align*}\)

Substitute each ordered pair to determine which makes the most money.

\(\begin{align*}(0,0) & \rightarrow \$0.00\\ (0,10) & \rightarrow 14.99(0)+16.99(10)=\$169.90\\ (11,0) &\rightarrow 14.99(11)+16.99(0)=\$164.89\\ (8,6) & \rightarrow 14.99(8)+16.99(6)=\$221.86\end{align*}\)

To make the most revenue, James should make 8 personal cakes and 6 cupcake orders.