Question

Consider the following linear program.

maximize 5x+3y

$$\begin{gathered} \mathbf{5}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\ge }\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{x}\mathbf{+}\mathit{y}\mathbf{\le }\mathbf{7} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{x}\mathbf{\le }\mathbf{5} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{x}\mathbf{\ge }\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\ge }\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Plot the feasible region and identify the optimal solution.

Short answer

The maximum value is 31 at (5,2) is the optimal solution and the feasible region is plotted.

Step-by-step solution

Explain feasible region and optimal solution

The feasible region is the region where the optimal solutions can be found between the axis. Optimal solution is the most appropriate solution of the problem.

Plot the feasible region and find the optimal solution

Consider that the obj (x,y).=5x+3y Assume that the x is 0, and y is 0, then,

obj (0,0) =0 .

Knowing that on x-axis ,y is 0,

$$\begin{gathered} obj\left(5,0\right)=5\times 5+3\times 0 \\ obj\left(5,0\right)=25+0 \\ obj\left(5,0\right)=25 \\ \mathrm{Since},\begin{array}{c}\mathrm{x}+\mathrm{y}\le 7\\ \mathrm{x}\le 5\end{array},\mathrm{Consider}\mathrm{x}=2,\mathrm{y}=5 \\ \mathrm{obj}\left(2,5\right)=2\times 5+3\times 5 \\ \mathrm{obj}\left(2,5\right)=10+15 \\ \mathrm{obj}\left(2,5\right)=25 \\ \mathrm{Now},\mathrm{Consider}\mathrm{x}=5,\mathrm{y}=2 \\ \mathrm{obj}\left(5,2\right)=5\times 5+3\times 2 \\ \mathrm{obj}\left(5,2\right)=25+6 \\ \mathrm{obj}\left(5,2\right)=31 \end{gathered}$$

The feasible region is plotted as follows

Therefore, the maximum value is 31 at (5,2) is the optimal solution and the feasible region is ploted.