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#### liang123993

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- Aug 12, 2019

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- Feb 5, 2013

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Let $W$ be width, $L$ be length an $A$ be the desired area. Then,

\(\displaystyle 5W+2L=550\)

\(\displaystyle LW=A\)

Can you make any progress from there?

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- Feb 5, 2013

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\(\displaystyle W=\frac AL\)

Let $W$ be width, $L$ be length an $A$ be the desired area. Then,

\(\displaystyle 5W+2L=550\)

\(\displaystyle LW=A\)

Can you make any progress from there?

\(\displaystyle \frac{5A}{L}+2L=550\)

\(\displaystyle 5A+2L^2=550L\)

\(\displaystyle A=110L-\frac{2L^2}{5}\)

$A$ has a maximum at the vertex of this inverted parabola, so $L=\frac{275}{2}$. Finding $A$ and $W$ from here should be straightforward.