A company produces two products, A and B. The sales volume for A is at least 80% of the total sales of both A and B. However, the company cannot sell more than 110 units of A per day. Both products use one raw material, of which the maximum daily availability is 300 lb. The usage rates of the raw material are 2 lb per unit of A, and 4 lb per unit of B. The profit units for A and B are $40 and $90, respectively. Determine the optimal product mix for the company.

Per pound of raw material, B produces 90/4 = $22.50 in profit, while A produces $40/2 = $20 in profit. Clearly, B is more profitable, so should be produced in the maximum possible number. Of course, the limitation is ultimately the amount of available raw material.

The requirement that A be at least 80% of the product mix means that at least 4 A must be produced for each B. Then, for each B produced, we have a raw material utilization of ...

4A(2 lb/A) +1B(4 lb/B) = 12 lb

The maximum number of B that can be produced is 300 lb/(12 lb/B) = 25 B.

Production of 25 B and 100 A is the optimal product mix.

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100 A is below the maximum of 110 that can be sold, so that is not a limit in this scenario.

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The attachment shows a graphical solution to the inequalities ...

x ≥ 0.80(x +y) ⇒ x ≥ 4y
x ≤ 110
2x + 4y ≤ 300
x ≥ 0, y ≥ 0

Quantities of A and B are represented by x and y, respectively.

The objective function 40x+90y is maximized when the line is at the vertex of the feasible region that puts it farthest from the origin. That point is (x, y) = (100, 25). A close candidate is the point (x, y) = (110, 20), for which the profit is slightly less.