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S = Total number of Standard Bags produced D= Total number of Deluxe Bags produced Profit 10 S + 9 D (Profit per standard bag is $9 and Delux bag is 10$) Max 10S + 9D s.t. 7/10S +1D ≤ 640 (Cutting and Dyeing) 1/2 S +5/6D ≤ 600 (Sewing) 1S + 2/3 D ≤ 708 (Finishing) 1/10S + 1/4D ≤135 (Inspection and Packaging) S ≥ 300 (You have to make 300 minimum standard bags) S+D ≥ 550 ( You have to make minimum of 550 total bags)

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Therefore, the optimal solution for the given problem is S = 300 and D = 250. The maximum profit is $5,200.

The given linear programming problem can be expressed as Maximize: Z = 10S + 9DSubject to:7/10S + D ≤ 6401/2S + 5/6D ≤ 600S + 2/3D ≤ 7081/10S + 1/4D ≤ 135S ≥ 300S + D ≥ 550

Where S = Total number of Standard Bags produced D = Total number of Deluxe Bags produced.

Profit per standard bag is $9 and for the deluxe bag is $10.Therefore, Maximize Z = 10S + 9D

The constraints of the problem are:7/10S + D ≤ 6401/2S + 5/6D ≤ 600S + 2/3D ≤ 7081/10S + 1/4D ≤ 135S ≥ 300S + D ≥ 550

Maximize Z = 10S + 9D

Subject to:7/10S + D ≤ 6401/2S + 5/6D ≤ 600S + 2/3D ≤ 7081/10S + 1/4D ≤ 135S ≥ 300S + D ≥ 550

To solve the problem, we can use the Simplex method.

The initial simplex tableau for the problem is:

| 10 | 9 | 0 | 0 | 0 | 0 | 0 ||---|---|---|---|---|---|---|| 7/10 | 1 | 1 | 0 | 0 | 0 | 640 || 1/2 | 5/6 | 0 | 1 | 0 | 0 | 600 || 1 | 2/3 | 0 | 0 | 1 | 0 | 708 || 1/10 | 1/4 | 0 | 0 | 0 | 1 | 135 || -1 | 0 | 0 | 0 | 0 | 0 | -300 || -1 | -1 | 0 | 0 | 0 | 0 | -550 |The first pivot element is 7/10.

Divide the first row by 7/10 to get 1 in the first element of the first row.

| 14 2/5 | 0.8571 | 0 | 0 | 0 | 0 | 915.43 ||----|-----|---|---|---|---|----|| 1 | 1.4 | 1 | 0 | 0 | 0 | 914.28 || 0 | 1.5 | -0.5 | 1 | 0 | 0 | 60 || 0 | 0.6667 | -0.6667 | 0 | 1 | 0 | 234 || 0 | 0.1 | -0.9 | 0 | 0 | 1 | 72 || 0 | -0.4 | 0 | 0 | 0 | 0 | -15 || 0 | -1 | 0 | 0 | 0 | 0 | -550 |

The next pivot element is 1.5.

Divide the second row by 1.5 to get 1 in the second element of the second row

.| 14 2/5 | 0.8571 | 0 | 0 | 0 | 0 | 915.43 ||----|-----|---|---|---|---|----|| 1 | 0.9333 | 0.6667 | 0 | 0 | 0 | 609.52 || 0 | 1 | -0.3333 | 0.6667 | 0 | 0 | 40 || 0 | 0.6667 | -0.6667 | 0 | 1 | 0 | 234 || 0 | 0.1 | -0.9 | 0 | 0 | 1 | 72 || 0 | -0.4 | 0 | 0 | 0 | 0 | -15 || 0 | -0.6667 | 0.3333 | 0 | 0 | 0 | -496.19 |The last pivot element is 0.6667.

Divide the fourth row by 0.6667 to get 1 in the second element of the fourth row.

| 15.857 | 0 | 1 | 0 | 0 | -7.143 | 769.52 ||----|----|---|---|---|------|------|| 0.3333 | 0 | 1.3333 | 0 | 0 | -0.6667 | 152.38 || 0 | 1 | -0.3333 | 0.6667 | 0 | 0 | 40 || 0 | 1 | -1 | 0 | 1.5 | 0 | 351.43 || 0 | 0.15 | -1.35 | 0 | 1.5 | 1 | 144 || 0 | 0 | 0.6667 | 0 | 0.4 | 0 | 6.429 || 0 | 0 | 1.3333 | 0 | 0.6667 | 0 | -345.71 |

The optimal solution is obtained when S = 300 and D = 250, and the maximum profit is $5,200. Hence, the company should produce 300 standard bags and 250 deluxe bags to maximize its profit.

Therefore, the optimal solution for the given problem is S = 300 and D = 250. The maximum profit is $5,200.

User Barunsthakur
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