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### Section 4-4 : Finding Absolute Extrema

10. Determine the absolute extrema of $$A\left( t \right) = {t^2}\,{\left( {10 - t} \right)^{\frac{2}{3}}}$$ on $$\left[ {2,\,\,10.5} \right]$$.

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Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem!
Start Solution

First, notice that we are working with a product of a polynomial and a cube root function. Both are continuous everywhere and so the product will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem!

Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function.

Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section.

Here are the critical points for this function.

\begin{align*}A'\left( t \right) & = 2t\,{\left( {10 - t} \right)^{\frac{2}{3}}} + {t^2}\left( {{\frac{2}{3}}} \right)\left( { - 1} \right)\,{\left( {10 - t} \right)^{ - \,\,\frac{1}{3}}} = 2t\,{\left( {10 - t} \right)^{\frac{2}{3}}} - \frac{{2{t^2}}}{{3{{\left( {10 - t} \right)}^{\frac{1}{3}}}}}\\ & = \frac{{6t\,\left( {10 - t} \right) - 2{t^2}}}{{3{{\left( {10 - t} \right)}^{\frac{1}{3}}}}} = \frac{{60t\, - 8{t^2}}}{{3{{\left( {10 - t} \right)}^{\frac{1}{3}}}}} = \frac{{4t\left( {15 - 2t} \right)}}{{3{{\left( {10 - t} \right)}^{\frac{1}{3}}}}}\\ & = 0\hspace{2.0in}t = 0,\,\,\,\,\,t = {\frac{15}{2}},\,\,\,\,\,\,t = 10\end{align*}

Don’t forget about critical points where the derivative doesn’t exist!

Show Step 2

Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical points that we need are,

$t = {\frac{15}{2}},\,\,\,\,\,\,t = 10$ Show Step 3

The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations.

$A\left( 2 \right) = 16\,\,\,\,\,\,\,A\left( {{\frac{15}{2}}} \right) = 103.613\,\,\,\,\,\,\,\,A\left( {10} \right) = 0\,\,\,\,\,\,\,\,\,A\left( {10.5} \right) = 69.4531$ Show Step 4

The final step is to identify the absolute extrema. So, the answers for this problem are then,

\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}{\mbox{Absolute Maximum : }} & 103.613{\mbox{ at }}t = {\frac{15}{2}}\\ {\mbox{Absolute Minimum : }} & 0{\mbox{ at }}t = 10\end{align*}}

Note the importance of paying attention to the interval with this problem. Had we neglected to exclude $$t = 0$$ we would have had the absolute minimum showing up at two places instead of only the one place inside the given interval.