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

The LPP will be solved graphically.
Step 1: Convert constraints to equalities and determine intercepts.
For the constraint $2x + y = 18$:
If $x = 0$, then $y = 18$.
If $y = 0$, then $x = 9$.
For the constraint $3x + 2y = 34$:
If $x = 0$, then $y = 17$.
If $y = 0$, then $x = \dfrac{34}{3} \approx 11.33$.
Next, find the intersection point of the two lines.
Solve the system: $2x + y = 18$ and $3x + 2y = 34$.
From the first equation, isolate $y$: $y = 18 - 2x$.
Substitute this into the second equation: $3x + 2(18 - 2x) = 34$.
Simplifying: $3x + 36 - 4x = 34$, which yields $-x = -2$, so $x = 2$.
Substitute $x = 2$ back into $y = 18 - 2x$: $y = 18 - 2(2) = 14$.
The intersection point is $(2,14)$.
Now, evaluate the objective function $z = 50x + 30y$ at all corner points of the feasible region.
At $(0,0)$: $z = 0$.
At $(0,17)$: $z = 0 + 510 = 510$.
At $(2,14)$: $z = 100 + 420 = 520$.
At $(9,0)$: $z = 450 + 0 = 450$.
The maximum value of $z$ is $520$, occurring at the point $(2,14)$.