• We revisit one of the most important applications of differential calculus: optimization.
• The problems we consider will involve exponential and trigonometric modelling functions.

The net monthly profit, in dollars, from the sale of a certain item is given by the formula \(P(x) = 60x^{2}e^{-0.25x}+40 \), where \(x \geq 1\) is the number of items sold, in hundreds.

a. If the factory can produce up to \(1000\) items per month, determine the number of items that yield the maximum profit. What is this maximum profit?

Solution

Since \(x=10\) corresponds to \(10\times 100 = 1000\) units of the item, we need to determine the maximum value of the profit function \(P(x)\) for \(1 \leq x \leq 10\). Since \(P(x)\) is continuous over all of \(\mathbb{R}\), it is continuous over the interval \([1,10]\).

By the extreme value theorem, \(P(x)\) must attain a maximum value on the interval \([1,10]\).

Let's locate the critical points of \(P(x)\) over this interval.

Using the product rule and the chain rule, we differentiate \(P(x)\):

We have

The net monthly profit, in dollars, from the sale of a certain item is given by the formula \(P(x) = 60x^{2}e^{-0.25x}+40 \), where \(x \geq 1\) is the number of items sold, in hundreds.

a. If the factory can produce up to \(1000\) items per month, determine the number of items that yield the maximum profit. What is this maximum profit?

Solution

Since \(e^{-0.25x} \neq 0\) for all \(x\), we have \(P'(x) = 0\) if and only if \(x=0\) or \(x=8\).

Therefore, \(P(x)\) has two critical points, but only one, \(x=8\), lies in the interval \([1,10]\).

We evaluate the function at the critical point and the endpoints of the interval to locate the maximum value of \(P(x)\):

Therefore, \(P(x)\) attains its maximum value at \(x=8\).

We conclude that the maximum profit attainable is \($559.69\), which is obtained by selling \(800\) items.

The net monthly profit, in dollars, from the sale of a certain item is given by the formula \(P(x) = 60x^{2}e^{-0.25x}+40 \), where \(x \geq 1\) is the number of items sold, in hundreds.

b. Repeat part a) assuming that the factory can produce only \(500\) items per month at full capacity.

Solution

Now we need to determine the maximum value of the profit function, \(P(x)\), for \(1 \leq x \leq 5\).

From part a), we know that \(P(x)\) has no critical points on this interval and so the maximum value must occur at one of the endpoints.

We have \(P(1) = 86.73\) and \(P(5) \approx 469.76\) and so \(P(x)\) attains its maximal value at \(x=5\), corresponding to \(500\) items.

The maximum profit is \($469.76\).