Application of derivatives: Profit analysis

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The sales revenue (in dollars) that a manufacturer receives for selling x units of a certain product can be approximated by the function

R(x) = 900 loge(1 + x/300)

Further, each unit costs the manufacturer one dollar to produce, and the initial cost of adjusting his machinery for production is $200, so that the total cost of production (in dollars) of x units is C(x) = 200 + x.

Write down the profit, P(x) dollars, obtained by the production and sale of x units and hence find the number of units which should be produced and sold for maximum profit. Calculate the resulting maximum profit.

I get that the P(x) is R(X) - C(x) = 900 loge (1 + x/300 ) − 200 − x

How do I differentiate it further?

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Simply use the chain rule and recall $(\ln x)’=\frac 1x$: $$P’(x)= 900\cdot \frac{1}{1+\frac{x}{300}} \cdot\frac{1}{300} -0-1 $$ Setting this equal to zero you’ll get $x=600$, which you can check by the second derivative test to correspond to a maximum.