Question

Maximize Z = 5x + 3y
Subject to

\[3x + 5y \leq 15\]
\[5x + 2y \leq 10\]
\[ x, y \geq 0\]

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:

3x + 5y = 15, 5x + 2y = 10, x = 0 and y = 0
Region represented by 3x + 5y ≤ 15 :
The line 3x + 5y = 15 meets the coordinate axes at A(5,0) and B(0,3) respectively. By joining these points we obtain the line 3x + 5y = 15.
Clearly (0,0) satisfies the inequation 3x + 5y ≤ 15. So,the region containing the origin represents the solution set of the inequation 3x + 5y ≤ 15.
Region represented by 5x + 2y ≤ 10 :
The line 5x + 2y = 10 meets the coordinate axes at C(2,0) and D(0,5) respectively. By joining these points we obtain the line 5x + 2y = 10.
Clearly (0,0) satisfies the inequation 5x + 2y ≤ 10. So,the region containing the origin represents the solution set of the inequation 5x + 2y ≤ 10.
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, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), C(2, 0),

\[E\left( \frac{20}{19}, \frac{45}{19} \right)\] and B(0, 3).

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

Corner point	Z = 5x + 3y
O(0, 0)	5 × 0 + 3 × 0 = 0
C(2, 0)	5 × 2 + 3 × 0 = 10
\[E\left( \frac{20}{19}, \frac{45}{19} \right)\]	5x \[\frac{20}{19}\] +3x \[\frac{45}{19}\] = \[\frac{235}{19}\]
B(0, 3)	5  0 + 3 × 3 = 9

 

Therefore, the maximum value of Z is

\[\frac{235}{19}\text{ at the point } \left( \frac{20}{19}, \frac{45}{19} \right)\] .Hence, x= \[\frac{20}{19}\]  and y =

\[\frac{45}{19}\]  is the optimal solution of the given LPP.
Thus, the optimal value of Z is \[\frac{235}{19}\] .