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A rectangle is formed by bending a length of wire of length $L$ around four pegs. Calculate the area of the largest rectangle which can be formed this way (as a function of $L$).

How am I supposed to do this? If I'm interpreting the question correctly, a square would have an area of $\dfrac{1}{16}L^2$. But I don't know how to find the maximum area. I'm guessing it involves finding the stationary point of some function of $L$, but which function that might be eludes me at the moment.

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If the rectangle is $h$ by $w$, we have the area is $A=wh$ and we have $2w+2h=L$. You solve the constraint to get $w=\frac 12(L-2h)$, and plug that into the other to get $A=\frac 12h(L-2h)$. Now take $\frac {dA}{dh}$ set it to zero, solve for $h$ and you are there. You will get the result you guessed.

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  • $\begingroup$ why do you take the derivative with respect to $h$ and not $L$? $\endgroup$ – Phaptitude Oct 16 '13 at 18:16
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    $\begingroup$ Because $L$ is a constant, $h$ is a variable. We were given $L$ and told to find $h$ $\endgroup$ – Ross Millikan Oct 16 '13 at 18:20
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Say length is $x$ and breadth is $y$ so you have got $2(x+y)=L$ now with respect to this you have to maximize $A=xy$

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You will get a rectangle of sides $a$ and $b$, whose area is $A=a\cdot b$, and perimeter $L=2a+2b$.

One approach is calculus: Let $x=a$, then $b=\frac L2-x$ and area is $A=x(\frac L2-x)=\frac L2x-x^2$, have the derivative equal $0$, and voila.

Second approach: $(a-b)^2\ge 0$ (equality only holds when $a-b=0$, so: \begin{align} a^2-2ab+b^2&\ge 0\\ 4ab+a^2-2ab+b^2&\ge 4ab \\ a^2+2ab+b^2&\ge 4ab \\ (a+b)^2&\ge 4ab \end{align}

Now, remeber $A=ab$ and $L=2a+2b$, hence $a+b=\frac L2$, and substituting: \begin{align} \left(\frac L2\right)^2 &\ge 4A \\ \frac{L^2}4&\ge4A \\ A&\le\frac{L^2}{16} \end{align}

The equality only holds when $a-b=0$ which is when $a=b=\frac L4$.

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