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Section 4-8 : Optimization

1. Find two positive numbers whose sum is 300 and whose product is a maximum.

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Start Solution

The first step is to write down equations describing this situation.

Let’s call the two numbers \(x\) and \(y\) and we are told that the sum is 300 (this is the constraint for the problem) or,

\[x + y = 300\]

We are being asked to maximize the product,

\[A = xy\] Show Step 2

We now need to solve the constraint for \(x\) or \(y\) (and it really doesn’t matter which variable we solve for in this case) and plug this into the product equation.

\[y = 300 - x\hspace{0.5in} \Rightarrow \hspace{0.5in}A\left( x \right) = x\left( {300 - x} \right) = 300x - {x^2}\] Show Step 3
The next step is to determine the critical points for this equation.

\[A'\left( x \right) = 300 - 2x\hspace{0.5in} \to \hspace{0.5in}300 - 2x = 0\hspace{0.5in} \to \hspace{0.5in}x = 150\] Show Step 4

Now for the step many neglect as unnecessary. Just because we got a single value we can’t just assume that this will give a maximum product. We need to do a quick check to see if it does give a maximum.

As discussed in notes there are several methods for doing this, but in this case we can quickly see that,

\[A''\left( x \right) = - 2\] Show Step 5

From this we can see that the second derivative is always negative and so \(A\left( x \right)\) will always be concave down and so the single critical point we got in Step 3 must be a relative maximum and hence must be the value that gives a maximum product.

Step 5

Finally, let’s actually answer the question. We need to give both values. We already have \(x\) so we need to determine \(y\) and that is easy to do from the constraint.

\[y = 300 - 150 = 150\]

The final answer is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = 150\hspace{0.5in}y = 150}}\]