The net monthly profit, in dollars, from the sale of a certain item, is given by the formula P(x)...

Question:

The net monthly profit, in dollars, from the sale of a certain item, is given by the formula {eq}P(x) = 10^6\left [ 1 + (x-1)e^{0.001x} \right ], {/eq} where x is the number of items sold. Find the number of items that yield the maximum profit. At full capacity, the factory can produce 2000 items per month.

Finding Maximum Profit from Sales of Item


The net monthly profit in dollars from the sale of a certain item is given as an explicit function of the number of items sold. Using a first derivative and second derivative test from Calculus, if applicable, we find the number of items sold that lead to a maximum value for the profit. The profit function contains the product of a linear and exponential term. The first derivative of such a function requires the use of the product rule for derivatives. Using the first and second derivative is an excellent tool used to find points of local maximum and minimum of a function.


Answer and Explanation:


To find the point of maximum of the profit P(x) we need to find its critical points. To that end we calculate its first derivative using the product rule for derivatives from Calculus as follows:

{eq}P'(x) = 10^6 \left[ 0 + (x-1)(0.001)e^{0.001x} + e^{0.001x}(1) \right] \qquad (1) {/eq}

Since P'(x) exists for all values of x, we find critical points of P(x) by solving P'(x) = 0 from (1) to get

{eq}10^6 \left[ 0 + (x-1)(0.001)e^{0.001x} + e^{0.001x}(1) \right] = 0 \implies e^{0.001x} \left[ 0.001(x-1)+1 \right] = 0 \qquad (2) {/eq}

From (2) we get the equation

{eq}0.001(x-1)+1 = 0 \implies 1 = (1-x)(0.001) \implies 1000 = 1-x \implies x = -999. {/eq}

Since a negative number of items is not possible, therefore no meaningful critical points exist. Hence a local maximum of the profit function P(x) does not exist.


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Solving Min-Max Problems Using Derivatives

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 1
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