You have 40m of fencing with which to enclose a rectangular space for a garden.find the largest area that can be enclosed with this much fencing and the dimensions of the corresponding garden.
Let the dimension for the space be $\displaystyle l$ by $\displaystyle w$, where $\displaystyle l$ denotes length and $\displaystyle w$ denotes width.
From given we have $\displaystyle 2l+2w=40$, also the area can be expressed as $\displaystyle A=l\times w$. Our goal is to maximize the area $\displaystyle A$.
Using the given perimeter constraint, we can write $\displaystyle l=20-w$. Then let's rewrite the area as: $\displaystyle A=(20-w)w=-w^2+20w$
Now the area is expressed as a function of $\displaystyle w$ and it is a quadratic function whose graph is a parabola open down. Hence it will have a maximum. More specific, we have
$\displaystyle A=-w^2+20w=-(w-10)^2+100$
which implies that when $\displaystyle w=10$, we obtain maximum area $\displaystyle A=100$. Thus the dimension should be $\displaystyle l=w=10$.