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Section 4.14 : Business Applications

1. A company can produce a maximum of 1500 widgets in a year. If they sell \(x\) widgets during the year then their profit, in dollars, is given by,

\[P\left( x \right) = 30,000,000 - 360,000x + 750{x^2} - \frac{1}{3}{x^3}\]

How many widgets should they try to sell in order to maximize their profit?

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Because these are essentially the same type of problems that we did in the Absolute Extrema section we will not be doing a lot of explanation to the steps here. If you need some practice on absolute extrema problems you should check out some of the examples and/or practice problems there.

All we really need to do here is determine the absolute maximum of the profit function and the value of \(x\) that will give the absolute maximum.

Here is the derivative of the profit function and the critical point(s) since we’ll need those for this problem.

\[P'\left( x \right) = - 360,000 + 1500x - {x^2} = - \left( {x - 1200} \right)\left( {x - 300} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = 300,\,\,\,x = 1200\] Show Step 2

From the problem statement we can see that we only want critical points that are in the interval \(\left[ {0,1500} \right]\). As we can see both of the critical points from the above step are in this interval and so we’ll need both of them.

Show Step 3

The next step is to evaluate the profit function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations.

\[\begin{align*}P\left( 0 \right) & = 30,000,000 & \hspace{0.75in} & P\left( {300} \right) = - 19,500,000\\ P\left( {1200} \right) & = 102,000,000 & \hspace{0.5in} & P\left( {1500} \right) = 52,500,000\end{align*}\] Show Step 4

From these evaluations we can see that they will need to sell 1200 widgets to maximize the profits.