Find maximum profit

derivativesoptimization

The cost of producing appliances is $C(x) = 300 + 2.2x^2$ where $x$ is the number produced per week.

If they are sold for $110 each, how many should be produced for maximum profit?

Solution: 25

Approach:

Profit = Revenue – Cost

$P(x) = R(x) – C(x)$

Let the cost of producing x number of units per week be $C(x) = 300 + 2.2x^2$ and the revenue $R(x) = 110x$ for selling x number of units.

Hence profit is $P(x) = 110x – 300 – 2.2x^2$.

To find the maximum profit, the derivative of $P(x) = 0$.

$P'(x) = 110 – 2.2x$

$110 – 2.2x = 0$

$x = 50$

The method above gives me x = 50 where the solution says 25. Is my understanding and approach correct?

Best Answer

your derivative is wrong. 𝑃′(𝑥)=110−4.4𝑥

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