Applied optimization

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Now we put our optimization skills to work.

In this section, we will present several worked examples of optimization problems. Our method for solving these problems is essentially the following:

Draw a picture.: If possible, draw a schematic picture with all the relevant information.
Determine your goal.: We need identify what needs to be optimized.
Find constraints.: What limitations are set on our optimization?
Solve for a single variable.: Now you should have a function to optimize.
Use calculus to find the extreme values.: Be sure to check your answer!

You are making cylindrical containers to contain a given volume. Suppose that the top and bottom are made of a material that is $N$ times as expensive (cost per unit area) as the material used for the lateral side of the cylinder. Find (in terms of $N$ ) the ratio of height to base radius of the cylinder that minimizes the cost of making the containers.

First we draw a picture:

Letting $c$ represent the cost of the lateral side, we can write an expression for the cost of materials: $C = 2\pi crh+\answer [given]{2\pi r^2Nc}.$ Since we know that $V=\pi r^2h$ , we can use this relationship to eliminate $h$ (we could eliminate $r$ , but it’s a little easier if we eliminate $h$ , which appears in only one place in the above formula for cost). We find $\begin{align*} C(r)&=2c\pi r\frac{V}{\pi r^2}+2Nc\pi r^2\\ &=\frac{2cV}{r}+2Nc\pi r^2. \end{align*}$

We want to know the minimum value of this function when $r$ is in $(0,\infty )$ . Setting $C'(r)=\answer [given]{-2cV/r^2+4Nc\pi r} =0$ we find $r=\sqrt [3]{V/(2N\pi )}$ . Since $C''(r)=\answer [given]{4cV/r^3+4Nc\pi }$ is positive when $r$ is positive, there is a local minimum at the critical value, and hence a global minimum since there is only one critical value.

Finally, since $h=V/(\pi r^2)$ ,

$\begin{align*} \frac{h}{r}&=\frac{V}{\pi r^3}\\ &=\frac{V}{\pi(V/(2N\pi))}\\ &=\answer[given]{2N}, \end{align*}$

so the minimum cost occurs when the height $h$ is $2N$ times the radius. If, for example, there is no difference in the cost of materials, the height is twice the radius.

You want to sell a certain number $n$ of items in order to maximize your profit. Market research tells you that if you set the price at $ $1.50$ , you will be able to sell $5000$ items, and for every $10$ cents you lower the price below $ $1.50$ you will be able to sell another $1000$ items. Suppose that your fixed costs (‘‘start-up costs’’) total $ $2000$ , and the per item cost of production (‘‘marginal cost’’) is $ $0.50$ . Find the price to set per item and the number of items sold in order to maximize profit, and also determine the maximum profit you can get.

The first step is to convert the problem into a function maximization problem. The revenue for selling $n$ items at $x$ dollars is given by $r(x) = nx$ and the cost of producing $n$ items is given by $c(x) = 2000+0.5 n.$ However, from the problem we see that the number of items sold is itself a function of $x$ , $n(x) =5000+\frac {1000(1.5-x)}{0.10}$ So profit is give by: $\begin{align*} P(x) &= r(x) - c(x)\\ &= nx - (2000+0.5 n)\\ &=-10000x^2+25000x-12000. \end{align*}$

We want to know the maximum value of this function when $x$ is between 0 and $1.5$ . The derivative is $P'(x)=\answer [given]{-20000x+25000},$ which is zero when $x=1.25$ . Since $P''(x)=-20000<0$ , there must be a local maximum at $x=1.25$ , and since this is the only critical value it must be a global maximum as well. Alternately, we could compute $P(0)=-12000$ , $P(1.25)=3625$ , and $P(1.5)=3000$ and note that $P(1.25)$ is the maximum of these. Thus the maximum profit is $ $3625$ , attained when we set the price at $ $1.25$ and sell $7500$ items. We can confirm our results by looking at the graph of $y=P(x)$ : $\graph [xmin=0,xmax=3,ymin=-4000,ymax=6000]{y=-10000x^2+25000x-12000}$