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 3x + 2y ≤ 80 :
The line 3x + 2y = 80 meets the coordinate axes at A(\( \dfrac{80}{3}\), 0) and B(0, 40) respectively. By joining these points we obtain the line 3x + 2y = 80.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 80 . So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 80 .
Region represented by 2x + 3y ≤ 70:
The line 2x + 3y = 70 meets the coordinate axes at C(35, 0) and D(0,\( \dfrac{70}{3}\)) respectively. By joining these points we obtain the line 2x + 3y ≤ 70.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 70. So,the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 70.
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 + 2y ≤ 80, 2x + 3y ≤ 70, x ≥ 0, and y ≥ 0 are as follows.
The corner points of the feasible region are O(0, 0), A(803, 0), E(20, 10) and D(0,\( \dfrac{70}{3}\))
The values of Z at these corner points are as follows.
| Corner point | Z = 15x + 10y |
| O(0, 0) | 15 × 0 + 10 × 0 = 0 |
| \( A(\dfrac{80}{3}, 0)\) | \(15 × \dfrac{80}{3} + 10 × 0 = 400\) |
| E(20, 10) | 15 × 20 + 10 × 10 = 400 |
| \( D(0,\dfrac{70}{3})\) | \( 15 × 0 + 10 ×\dfrac{70}{3} = \dfrac{700}{3}\) |
We see that the maximum value of the objective function Z is 400 which is at A(\( \dfrac{80}{3}\), 0) and E(20, 10)
Thus, the optimal value of Z is 400.
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x≥3, y≥2
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subject to
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