Question

In a certain chemical manufacturing process, the daily weight y of defective chemical output depends on the total weight.x of all output according to the empirical formula y = 0.02x +0.0005x² where x and y are in pounds. If the profit is $300 per pound of non-defective chemical produced and the loss is $60 per pound of defective chemical produced, how many pounds of chemical should be produced daily to maximize the total daily profit? Round your answer to the nearest integer. pounds maximizes the total daily profit.

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Answer 1

813 pounds.

Explanation:

First write a formula for the profit:

Profit P = 300(x - 0.02x - 0.0005x²) - 60(0.02x + 0.0005x²)

We need to maximise this so we first find the derivative:

dP/dx = 300(1 - 0.02 - 0.001x) - 60(0.02 + 0.001x) = 0

for a maximum value)

300 - 6 - 0.3x - 1.2 - 0.06x = 0

- 0.36x = -292.8

x = 813.33