Question

Solve the linear programming problem graphically. Maximize: \( z = 3x + 5y \) Subject to: \[ x + 4y \leq 24, 3x + y \leq 21, x + y \leq 9, x \geq 0, y \geq 0 \] Also, find the maximum value of \( z \).

Hint:

In linear programming problems, graph the constraints to find the feasible region, then evaluate the objective function at each corner point to find the maximum value.

Answer 1

Step 1: Graph the constraints.
The first inequality is \( x + 4y \leq 24 \). Rewriting this as \( y \leq \frac{24 - x}{4} \), we plot the line \( x + 4y = 24 \). The second inequality is \( 3x + y \leq 21 \). Rewriting this as \( y \leq 21 - 3x \), we plot the line \( 3x + y = 21 \). The third inequality is \( x + y \leq 9 \). Rewriting this as \( y \leq 9 - x \), we plot the line \( x + y = 9 \). The constraints \( x \geq 0 \) and \( y \geq 0 \) mean that the feasible region is in the first quadrant.

Step 2: Find the feasible region.
The feasible region is the area where all the constraints are satisfied simultaneously. This region is bounded by the lines \( x + 4y = 24 \), \( 3x + y = 21 \), and \( x + y = 9 \), and is confined to the first quadrant.

Step 3: Find the corner points.
The corner points of the feasible region are found by solving the system of equations corresponding to the intersection points of the constraint lines. The corner points are: \[ (0, 0), (0, 6), (4, 5), (7, 2) \]

Step 4: Calculate the value of \( z \) at each corner point.
Substitute the coordinates of each corner point into the objective function \( z = 3x + 5y \): - At \( (0, 0) \): \( z = 3(0) + 5(0) = 0 \) - At \( (0, 6) \): \( z = 3(0) + 5(6) = 30 \) - At \( (4, 5) \): \( z = 3(4) + 5(5) = 12 + 25 = 37 \) - At \( (7, 2) \): \( z = 3(7) + 5(2) = 21 + 10 = 31 \)

Step 5: Find the maximum value of \( z \).
The maximum value of \( z \) occurs at \( (4, 5) \), where \( z = 37 \).

Final Answer: The maximum value of \( z \) is \( \boxed{37} \) at the point \( (4, 5) \).