Question
Maximum Z = 15x + 10y
Subject to
$3\text{x}+2\text{y}\leq80$
$2\text{x}+3\text{y}\leq70$
$\text{x},\text{y}\geq0$
Subject to
$3\text{x}+2\text{y}\leq80$
$2\text{x}+3\text{y}\leq70$
$\text{x},\text{y}\geq0$
✓
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 2y = 80, 2x + 3y = 70, x = 0 and y=0
Region represented by $3\text{x}+2\text{y}\leq80:$
The line 3x + 2y = 80 meets the coordinate axes at $\text{A}\Big(\frac{80}{3},0\Big)$ and B(0, 40) respectively.
By joining these points we obtain the line 3x + 2y = 80.
Clearly (0,0) satisfies the inequation $3\text{x}+2\text{y}\leq80$.
So, the region containing the origin represents the solution set of the inequation $3\text{x}+2\text{y}\leq80$.
Region represented by $2\text{x}+3\text{y}\leq70:$
The line 2x + 3y = 70 meets the coordinate axes at C(35, 0) and $\text{D}\Big(0,\frac{70}{3}\Big)$ respectively.
By joining these points we obtain the line $2\text{x}+3\text{y}\leq70$.
Clearly (0,0) satisfies the inequation $2\text{x}+3\text{y}\leq70$.
So, the region containing the origin represents the solution set of the inequation $2\text{x}+3\text{y}\leq70$.
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 $3\text{x}+2\text{y}\leq80$, $2\text{x}+3\text{y}\leq70$, $\text{x}\geq0$ and $\text{y}\geq0$ are as follows.
The corner points of the feasible are O(0, 0), $\text{A}\Big(\frac{80}{3},0\Big)\text{E}(20,10)$ and $\text{D}\Big(0,\frac{700}{3}\Big)$ .
The values of Z at these corner point are as follows.
We see that maximum value of the objective functioin Z is 400 which is at $\text{A}\Big(\frac{80}{3},0\Big)$ and E(20, 10).
Thus, the optimal value of Z is 400.
3x + 2y = 80, 2x + 3y = 70, x = 0 and y=0
Region represented by $3\text{x}+2\text{y}\leq80:$
The line 3x + 2y = 80 meets the coordinate axes at $\text{A}\Big(\frac{80}{3},0\Big)$ and B(0, 40) respectively.
By joining these points we obtain the line 3x + 2y = 80.
Clearly (0,0) satisfies the inequation $3\text{x}+2\text{y}\leq80$.
So, the region containing the origin represents the solution set of the inequation $3\text{x}+2\text{y}\leq80$.
Region represented by $2\text{x}+3\text{y}\leq70:$
The line 2x + 3y = 70 meets the coordinate axes at C(35, 0) and $\text{D}\Big(0,\frac{70}{3}\Big)$ respectively.
By joining these points we obtain the line $2\text{x}+3\text{y}\leq70$.
Clearly (0,0) satisfies the inequation $2\text{x}+3\text{y}\leq70$.
So, the region containing the origin represents the solution set of the inequation $2\text{x}+3\text{y}\leq70$.
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 $3\text{x}+2\text{y}\leq80$, $2\text{x}+3\text{y}\leq70$, $\text{x}\geq0$ and $\text{y}\geq0$ are as follows.
The corner points of the feasible are O(0, 0), $\text{A}\Big(\frac{80}{3},0\Big)\text{E}(20,10)$ and $\text{D}\Big(0,\frac{700}{3}\Big)$ .
The values of Z at these corner point are as follows.
|
$\text{Corner point}$
|
$\text{Z}=15\text{x}+10\text{y}$
|
|
$\text{O}(0, 0)$
|
$15\times0+10\times0=0$
|
|
$\text{A}\Big(\frac{80}{3},0\Big)$
|
$15\times\frac{80}{3}+10\times0=400$
|
|
$\text{E}(20, 10)$
|
$15\times20+10\times10=400$
|
|
$\text{D}\Big(0,\frac{70}{3}\Big)$
|
$15\times0+10\times\frac{70}{3}=\frac{700}{3}$
|
Thus, the optimal value of Z is 400.
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