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 SolutionIn 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 2The 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 3Now 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 4Finally, 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{{53}}{2}\hspace{0.5in}y = \frac{{47}}{4}}}\]