Question
Maximize Z = 5x + 3y
Subject to
$3\text{x}+5\text{y}\leq15$
$5\text{x}+2\text{y}\leq10$
$\text{x},\text{y}\geq0$
Subject to
$3\text{x}+5\text{y}\leq15$
$5\text{x}+2\text{y}\leq10$
$\text{x},\text{y}\geq0$
✓
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 5y = 15, 5x + 2 y = 10, x = 0 and y = 0
Region represented by $3\text{x}+5\text{y}\leq15:$
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 $3\text{x}+5\text{y}\leq15$.
So, the region containing the origin represents the solution set of the inequation $3\text{x}+5\text{y}\leq15$.
Region represented by $5\text{x}+2\text{y}\leq10:$
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 $5\text{x}+2\text{y}\leq10$.
So, the region containing the origin represents the solution set of the inequation $5\text{x}+2\text{y}\leq10$.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $\text{x}\geq0$, and $\text{y}\geq0$.
The feasible region determined by the system of constraints, $3\text{x}+5\text{y}\leq15,5\text{x}+2\text{y}\leq10,\text{x}\geq0$ and $\text{y}\geq0$,are as follows.
3x + 5y = 15, 5x + 2 y = 10, x = 0 and y = 0
Region represented by $3\text{x}+5\text{y}\leq15:$
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 $3\text{x}+5\text{y}\leq15$.
So, the region containing the origin represents the solution set of the inequation $3\text{x}+5\text{y}\leq15$.
Region represented by $5\text{x}+2\text{y}\leq10:$
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 $5\text{x}+2\text{y}\leq10$.
So, the region containing the origin represents the solution set of the inequation $5\text{x}+2\text{y}\leq10$.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $\text{x}\geq0$, and $\text{y}\geq0$.
The feasible region determined by the system of constraints, $3\text{x}+5\text{y}\leq15,5\text{x}+2\text{y}\leq10,\text{x}\geq0$ and $\text{y}\geq0$,are as follows.
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