Maximize Z = 10x + 6y

Subject to

3x+y≤122x+5y≤34  x, y≥0

Asked by Sakshi | 1 year ago |  172

##### Solution :-

First, we will convert the given inequations into equations, we obtain the following equations:
3x + y = 12, 2x + 5y = 34, x = 0 and y = 0

Region represented by 3x + y ≤ 12:
The line 3x + y = 12 meets the coordinate axes at A(4, 0)A4, 0 and B(0, 12)B0, 12 respectively. By joining these points we obtain the line 3x + y = 12.
Clearly (0,0) satisfies the inequation 3x + y ≤ 12. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 12 .

Region represented by 2x + 5y  ≤ 34:
The line 2x + 5y = 34 meets the coordinate axes at C(17, 0)C17, 0 and $$D(0,\dfrac{34}{5})$$ respectively. By joining these points we obtain the line 2x + 5y  ≤ 34.
Clearly (0,0) satisfies the inequation 2x + 5y  ≤ 34. So,the region containing the origin represents the solution set of the inequation 2x + 5y  ≤ 34.

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 + y ≤ 12, 2x + 5y  ≤ 34, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the feasible region are O(0, 0), A(4, 0)A4, 0 ,E(2, 6)E2, 6 and $$D(0,\dfrac{34}{5})$$

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

 Corner point Z = 10x + 6y O(0, 0) 10 × 0 + 6 × 0 = 0 A(4, 0) 10× 4 + 6 × 0 = 40 E(2, 6) 10 × 2 + 6 × 6 = 56 $$D(0,\dfrac{34}{5})$$ $$10 × 0 + 6 ×\dfrac{34}{5}=\dfrac{204}{3}$$

We see that the maximum value of the objective function Z is 56 which is at E(2, 6)E2, 6 that means at x = 2 and y = 6.
Thus, the optimal value of Z is 56.

Answered by Aaryan | 1 year ago

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