Q. 52 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 51 and 52, solve each linear programming problem.

Minimize $z=3x+5y$ when $x\ge 0; y\ge 0; 3x+2y\le 12; x+3y\le 12$

Step-by-Step Solution

Answer

By minimizing $z=3x+5y$ when $x\ge 0; y\ge 0; 3x+2y\le 12; x+3y\le 12$, the maximum value is $3$which occurs at $x=1; y=0$

Step 1. Given

The problem: 

$z=3x+5y$

The constraints: 

$$\begin{gathered} x\ge 0; \\ y\ge 0; \\ 3x+2y\le 12; \\ x+3y\le 12 \end{gathered}$$

Step 2. Graph the constraints.

The graph of the constraints (the feasible points) is illustrated:

Step 3. List the corner points

We list the corner points and evaluate the objective function at each.

S.No.	Corner points(x, y)	Value of the objective function $z=3x+5y$
1.	$(1,0)$	$$\begin{gathered} z=3\left(1\right)+5\left(0\right) \\ =3 \end{gathered}$$
2.	$(4,0)$	$$\begin{gathered} z=3\left(4\right)+5\left(0\right) \\ =12 \end{gathered}$$
3.	$(1.71, 3.43)$	$$\begin{gathered} z=3(1.71)+5(3.43) \\ =22.1 \end{gathered}$$
4.	$(0,4)$	$$\begin{gathered} z=3\left(0\right)+5\left(4\right) \\ =20 \end{gathered}$$
5.	$(0,1)$	$$\begin{gathered} z=3\left(0\right)+5\left(1\right) \\ =5 \end{gathered}$$

Step 4. Minimize the objective function.

From the table, the minimum value of the given objective function is $3$ and it occurs at $x=1; y=0$