Find the area of the largest rectangle 
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.
 A: 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 $A=xy$
A: 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.
A: 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 $a=b=\frac L4$.
