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\hfil Math 120 A, B - Winter 2010 \hfil \pp
\hfil Mid-Term Exam Number Two\hfil  \pp
\hfil February 25, 2010 \hfil \pp
\hfil Answers \hfil \pp
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There were two versions of the exam in use.

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Version A: Problem 1 involved 100 cm of wire.


\begin{enumerate}

\item (a) To minimize the combined area, the square should have side length
10.87411 cm, and the triangle should have side length 18.8345 cm.
(b) To maximize the combined area, the square should have length 25 cm, and the
triangle should have side length 0 cm.
\item (a) $\dsps f(x) = \frac{40x+145}{x+32.5}$ (b) -8.8620 and 16.3620
\item 37.1213 years after 1960
\item (a) $\dsps f(x) = \frac{3x}{-2x-3}$ (b) $\dsps f^{-1}(x) = \frac{-3x}{3+2x}$
\end{enumerate}

\bigskip
Version B: Problem 2 involved 80 cm of wire.

\begin{enumerate}
\item (a) To minimize the combined area, the square should have side length
8.69929 cm, and the triangle should have side length 15.0676 cm.
(b) To maximize the combined area, the square should have length 20 cm, and the
triangle should have side length 0 cm.
\item (a) $\dsps f(x) = \frac{30x+150}{x+20.5}$ (b) -8.386 and 17.886
\item 24.605 years after 1960
\item (a) $\dsps f(x) = \frac{2x}{8x-28}$ (b) $\dsps f^{-1}(x) = \frac{-28x}{2-8x}$
\end{enumerate}



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