A retail store sells two types of shoes, sneakers and sandals. The store owner pays $8 for the sneakers and $14 for the sandals. The sneakers can be sold for $10 and the sandals can be sold for $17. The owner of the store estimates that she won't sell more than 200 shoes each month, and doesn't plan to invest more that $2,000 on inventory of the shoes. If she sold the number of sneakers and sandals that could maximize her profit, what would her maximum profit be?

P = $

Question

A retail store sells two types of shoes, sneakers and sandals.  The store owner pays $8 for the sneakers and $14 for the sandals.  The sneakers can be sold for $10 and the sandals can be sold for $17.  The owner of the store estimates that she won't sell more than 200 shoes each month, and doesn't plan to invest more that $2,000 on inventory of the shoes.  If she sold the number of sneakers and sandals that could maximize her profit, what would her maximum profit be?

P = $

BlackTom AI · Accepted Answer

We need to maximize profit given two product types with constraints on quantity and inventory cost.

First, define variables: let x be the number of sneakers sold, and y be the number of sandals sold.
- Each sneaker yields a profit of selling price minus cost: 10 − 8 = 2 dollars per pair.
- Each sandal yields a profit of selling price minus cost: 17 − 14 = 3 dollars per pair.
- Total profit P = 2x + 3y.

Constraints:
- Quantity constraint: x + y ≤ 200 (she won’t sell more than 200 shoes per month).
- Inventory constraint: 8x + 14y ≤ 2000 (she won’t invest more than $2000 in inventory).
- Nonnegativity: x ≥ 0, y ≥ 0.

To find the maximum, we examine the feasible region defined by these constraints. The two main boundary lines are:
- x + y = 200
- 8x + 14y = 2000 (which can be simplified to 4x + 7y = 1000 after dividing by 2)

The optimal solution for a linear program with two variables typically occurs at a vertex (corner) of the feasible region or along an edge where the gradient aligns with a constraint.

Compute potential corner points:
- Intercepts: For x = 0, 4x + 7y ≤ 1000 gives y ≤ 1000/7 ≈ 142.86, and x + y ≤ 200 gives y ≤ 200. So the binding one is y ≤ 142, giving point (0,142) with profit P = 2(0) + 3(142) = 426.
- For y = 0, 4x ≤ 1000 gives x ≤ 250, and x + y ≤ 200 gives x ≤ 200. So the binding one is x ≤ 200, giving point (200,0) with profit P = 2(200) + 3(0) = 400.
- Intersection of the two constraint lines x + y = 200 and 4x + 7y = 1000 would be a corner, but solving these simultaneously gives a non-integer solution: x ≈ 133.33, y ≈ 66.67. In integer terms, we look for nearby integer pairs that satisfy both constraints.

Because the sandals have a higher per-unit profit (3 vs 2), the optimal integer solution tends to use as much of y as possible while staying within the constraints. Check integer pairs that satisfy 4x + 7y ≤ 1000 and x + y ≤ 200, focusing on larger y values and adjusting x accordingly.

A convenient way is to examine feasible pairs along the line 4x + 7y = 1000 with integer x values that also satisfy x + y ≤ 200. Sampling such pairs (where x ≡ 5 mod 7 to meet 4x + 7y = 1000 with integer y) yields:
- (x, y) = (131, 68) gives 4(131) + 7(68) = 524 + 476 = 1000 and x + y = 199 ≤ 200, so feasible. P = 2(131) + 3(68) = 262 + 204 = 466.
- Nearby integer pairs along this line tend to produce slightly smaller profits; for example (124, 72) yields P = 248 + 216 = 464, and (138, 64) yields P = 276 + 192 = 468 but check feasibility: 4(138) + 7(64) = 552 + 448 = 1000 exactly and x + y = 202, which violates the total quantity constraint, so it is not feasible.
Thus, the most favorable feasible integer combination appears to be at (131, 68), or nearby variants that respect both constraints and keep a high y value, yielding a higher overall profit than the other corners.

Putting it together, the maximum attainable profit under the given constraints arises from choosing the mix that uses the higher-margin sandal product as much as possible without violating inventory or total quantity limits, with the calculated feasible integer point around x = 131 sneakers and y = 68 sandals, which produces a total profit on the order of several hundred dollars. The exact value depends on ensuring both constraints are satisfied and using the highest-profit combination within that feasible region.

ALGII141-027 3.7 Practice Problems

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