Question

Solved the following linear programming problem graphically:
Maximize Z = 60x + 15y
Subject to constraints
$$\begin{gathered} x+y\le 50 \\ 3x+y\le 90 \\ x,y\ge 0 \end{gathered}$$

Answer 1

We have to maximize Z = 60x + 15y
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 50, 3x + y = 90, x = 0 and y = 0

Region represented by x + y ≤ 50:
The line x + y = 50 meets the coordinate axes at A(50,0) and B(0,50) respectively. By joining these points we obtain the line 3x + 5y = 15.
Clearly (0,0) satisfies the inequation x + y ≤ 50. So,the region containing the origin represents the solution set of the inequation x + y ≤ 50.

Region represented by 3x + y ≤ 90:
The line 3x + y = 90 meets the coordinate axes at C(30, 0) and D(0, 90) respectively. By joining these points we obtain the line 3x + y = 90.
Clearly (0,0) satisfies the inequation 3x + y ≤ 90. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 90.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints, x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), C(30, 0), $E\left(20,30\right)$ and B(0, 50).

The values of Z at these corner points are as follows.

Corner point	Z = 60x + 15y
O(0, 0)	60 × 0 + 15 × 0 = 0
C(30, 0)	60 × 30 + 15 × 0 = 1800
$E\left(20,30\right)$	60 × 20 + 15 × 30 =1650
B(0, 50)	60 × 0 + 15 × 50 = 750

Therefore, the maximum value of Z is $1800\mathrm{at}\mathrm{the}\mathrm{point}\left(30,0\right)$.Hence, x = 30 and y = 0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 1800.