Answer:
The optimal usage of fabric = 2
Step-by-step explanation:
Given the quantity, Q = 10 + 4F - (1/3) F^3
Selling price = $20
Profit = TR - TC
There is no variable cost and let the fixed cost is constant G.
Profit = PQ - G
Profit = 20(10 + 4F − (1/3)F^3)) - G = 0
Now take the first order derivative:
d(profit) / dF = 0
20(4 - F^2) = 0
F = 2
Therefore the optimal usage of fabric = 2