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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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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.

$y = 300 - 150 = 150$

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