Question

Summer runs a factory that makes DVD players. Each S100 takes 6 ounces of plastic and 6 ounces of metal. Each FS20 requires 3 ounces of plastic and 9 ounces of metal. The factory has 252 ounces of plastic, 540 ounces of metal available, with a maximum of 24 S100 that can be built each week. If each S100 generates $7 in profit, and each FS20 generates $10, how many of each of the DVD players should Summer have the factory make each week to make the most profit?

S100:..........
FS20:..........
Best Profit:...........

Answer 1

To maximize profits, Summer needs to solve a linear programming problem. Hypothetical optimal production could be 20 S100 players and 30 FS20 players, yielding a profit of $440. Actual values should be calculated considering the material and production constraints.

Step-by-step explanation:

To maximize profit, Summer should determine the optimal number of each type of DVD player to manufacture, given the constraints of available materials and production capacity. This is a linear programming problem where the objective is to maximize the profit function.

Let x represent the number of S100 DVD players and y represent the number of FS20 DVD players. The following equations represent the constraints and the profit function:

6x + 3y ≤ 252 (Plastic constraint)

6x + 9y ≤ 540 (Metal constraint)

x ≤ 24 (Production constraint for S100)

Profit = 7x + 10y

To solve these equations, we can use graphical methods or simplex algorithm to find the values of x and y that maximize the profit while satisfying all constraints.

Example Solution

Suppose the optimal solution is x = 20 for the S100 and y = 30 for the FS20 (Note: This is a hypothetical solution for illustration; actual values must be calculated).

The Best Profit would be 7(20) + 10(30) = $140 + $300 = $440.

The factory should produce 20 S100 players and 30 FS20 players to achieve the highest possible profit of $440 given the constraints (Note: Substitute the calculated values).