Question

Solve the following linear programming problem graphically:
Maximize $z = 50x + 30y$
Subject to:
$2x + y \leq 18$
$3x + 2y \leq 34$
$x \geq 0$, $y \geq 0$

Hint:

Plot the constraints, shade the feasible region, find vertices, and evaluate the objective function at each vertex to find maximum or minimum.

Answer 1

We are to solve the LPP using the graphical method.
Step 1: Convert constraints to equalities and find intercepts.
For $2x + y = 18$:
$x = 0 \Rightarrow y = 18$,
$y = 0 \Rightarrow x = 9$
For $3x + 2y = 34$:
$x = 0 \Rightarrow y = 17$,
$y = 0 \Rightarrow x = \dfrac{34}{3} \approx 11.33$
Now find the point of intersection of the two lines:
Solve: $2x + y = 18$ and $3x + 2y = 34$
From first: $y = 18 - 2x$
Substitute in second: $3x + 2(18 - 2x) = 34$
$3x + 36 - 4x = 34 \Rightarrow -x = -2 \Rightarrow x = 2$
Then $y = 18 - 2(2) = 14$
So point of intersection: $(2,14)$
Now evaluate $z = 50x + 30y$ at all corner points of feasible region:
At (0,0): $z = 0$
At (0,17): $z = 0 + 510 = 510$
At (2,14): $z = 100 + 420 = 520$
At (9,0): $z = 450 + 0 = 450$
Maximum value of $z$ occurs at $(2,14)$ with $z = 520$