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Section 4-8 : Optimization

3. Let \(x\) and \(y\) be two positive numbers such that \(x + 2y = 50\) and \(\left( {x + 1} \right)\left( {y + 2} \right)\) is a maximum.

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Start Solution

In this case we were given the constraint in the problem,

\[x + 2y = 50\]

We are also told the equation to maximize,

\[f = \left( {x + 1} \right)\left( {y + 2} \right)\]

So, let’s just solve the constraint for \(x\) or \(y\) (we’ll solve for \(x\) to avoid fractions…) and plug this into the product equation.

\[x = 50 - 2y\hspace{0.25in} \Rightarrow \hspace{0.25in}f\left( y \right) = \left( {50 - 2y + 1} \right)\left( {y + 2} \right) = \left( {51 - 2y} \right)\left( {y + 2} \right) = 102 + 47y - 2{y^2}\] Show Step 2

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

\[f'\left( y \right) = 47 - 4y\hspace{0.25in} \to \hspace{0.25in}47 - 4y = 0\hspace{0.5in} \to \hspace{0.5in}y = \frac{{47}}{4}\] Show Step 3

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,

\[f''\left( y \right) = - 4\]

From this we can see that the second derivative is always negative and so \(f\left( y \right)\) will always be concave down and so the single critical point we got in Step 2 must be a relative maximum and hence must be the value that gives a maximum.

Show Step 4

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.

\[x = 50 - 2\left( {\frac{{47}}{4}} \right) = \frac{{53}}{2}\]

The final answer is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{{53}}{2}\hspace{0.5in}y = \frac{{47}}{4}}}\]