In a certain chemical manufacturing process, the daily weight y of defective che

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 + 0.0007x^2
where x and y are in pounds. If the profit is dollar-sign 400 per pound of nondefective chemical produced and the loss is dollar-sign 80 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.

Explanation / Answer

Weight of defective part = y = 0.02x + 0.0007x2

Non defective part = x-y = x-y = 0.02x - 0.0007x2

profit for non defective part = 400x

loss for defective part = 80y = 80(0.02x + 0.0007x^2) = 1.60x+0.056x2

Net profit = 400x-1.60x-0.056x2

Find derivative for profit

P' = 400-1.60-0.112x

P'=0 gives

0.112x = 398.4

x = 3554.14

So for 3555 units profit is maximum. ( it is rounded off to 3555 as units cannot be in decimals, and it is safe to take the higher integer here)

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