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### Section 4.8 : Optimization

2. Find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum.

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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 product is 750 (this is the constraint for the problem) or,

$xy = 750$

We are then being asked to minimize the sum of one and 10 times the other,

$S = x + 10y$

Note that it really doesn’t worry which is $$x$$ and which is $$y$$ in the sum so we simply chose the $$y$$ to be multiplied by 10.

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.

$x = \frac{{750}}{y}\hspace{0.5in} \Rightarrow \hspace{0.5in}S\left( y \right) = \frac{{750}}{y} + 10y$ Show Step 3

The next step is to determine the critical points for this equation.

$S'\left( y \right) = - \frac{{750}}{{{y^2}}} + 10\hspace{0.5in} \to \hspace{0.5in} - \frac{{750}}{{{y^2}}} + 10 = 0\hspace{0.5in} \to \hspace{0.5in}y = \pm \sqrt {75} = 5\sqrt 3$

Because we are told that $$y$$ must be positive we can eliminate the negative value and so the only value we really get out of this step is : $$y = \sqrt {75} = 5\sqrt 3$$.

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 minimum sum. We need to do a quick check to see if it does give a minimum.

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

$S''\left( y \right) = \frac{{1500}}{{{y^3}}}$

From this we can see that, provided we recall that $$y$$ is positive, then the second derivative will always be positive. Therefore, $$S\left( y \right)$$ will always be concave up and so the single critical point from Step 3 that we can use must be a relative minimum and hence must be the value that gives a minimum sum.

Show Step 5

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

$x = \frac{{750}}{{5\sqrt 3 }} = 50\sqrt 3$

$\require{bbox} \bbox[2pt,border:1px solid black]{{x = 50\sqrt 3 \hspace{0.5in}y = 5\sqrt 3 }}$