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Update singVarCalcOptimization3.tex #976

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15 changes: 6 additions & 9 deletions optimization/exercises/singVarCalcOptimization3.tex
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Expand Up @@ -34,11 +34,11 @@
\end{tikzpicture}
\end{hint}
\begin{hint}
The dimensions that will result in "least material" for the five sides are dimensions that result in "smallest surface area".
So, the quantity that we have to "minimize" is the surface area, $S$.
The dimensions that will result in ``least material'' for the five sides are dimensions that result in ``smallest surface area''.
So, the quantity that we have to ``minimize'' is the surface area, $S$.
\end{hint}
\begin{hint}
We have to "minimize" the surface area, $S$.
We have to ``minimize'' the surface area, $S$.

$S=x^2 +4xy$

Expand All @@ -64,9 +64,8 @@
\begin{hint}
So, we have to find the (global) minimum of $S$ on its domain, $(0,\answer{\infty})$.
We have to find critical points of $S$.
Therefore, we have to compute $S'(x)$.
\end{hint}
\begin{hint}
Therefore, we have to compute $S'(x)$:

$S'(x)=2x-\frac{400}{\answer{x^2}}$.
\end{hint}
\begin{hint}
Expand All @@ -76,9 +75,7 @@
$2x-\frac{400}{\answer{x^2}}=0$.

It follows that $S$ has the only critical point $x=\answer{200^{\frac{1}{3}}}$.
\end{hint}

\begin{hint}

Since

$S'(x)=2x-\frac{400}{x^{2}}=\frac{2(x^{3}-200)}{x^{2}}$, it follows that
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