Section 4.8 : Optimization
3. Let x and y be two positive numbers such that x+2y=50 and (x+1)(y+2) is a maximum.
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Start SolutionIn this case we were given the constraint in the problem,
x+2y=50We are also told the equation to maximize,
f=(x+1)(y+2)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⇒f(y)=(50−2y+1)(y+2)=(51−2y)(y+2)=102+47y−2y2 Show Step 2The next step is to determine the critical points for this equation.
f′(y)=47−4y→47−4y=0→y=474 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″(y)=−4From this we can see that the second derivative is always negative and so f(y) 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(474)=532The final answer is then,
x=532y=474