Originally Posted by
Chris L T521 Let a rectangle have length $\displaystyle l$ and width $\displaystyle w$.
Then by the first condition, we have $\displaystyle P=2l+2w=80$.
Let $\displaystyle A=lw$. It follows from above that $\displaystyle l=40-w$.
Substituting this into the area equation, we have $\displaystyle A=(40-w)w=40w-w^2$.
Since we want to maximize area, it follows that $\displaystyle \frac{\,dA}{\,dw}=40-2w$. Now, we find the critical point:
$\displaystyle 40-2w=0\implies w=20$ (To verify its a max, we differentiate again to get $\displaystyle A^{\prime\prime}=-2<0$)
Since $\displaystyle w=20$, it follows that $\displaystyle l=40-(20)=20$.
Therefore, the dimensions of the rectangle that gives us the maximum area actually form a square.
Does this make sense?