maximize problem

Let the dimension for the space be $l$ by $w$ , where $l$ denotes length and $w$ denotes width.

From given we have $2l+2w=40$ , also the area can be expressed as $A=l\times w$ . Our goal is to maximize the area $A$ .

Using the given perimeter constraint, we can write $l=20-w$ . Then let's rewrite the area as: $A=(20-w)w=-w^2+20w$

Now the area is expressed as a function of $w$ and it is a quadratic function whose graph is a parabola open down. Hence it will have a maximum. More specific, we have

$A=-w^2+20w=-(w-10)^2+100$

which implies that when $w=10$ , we obtain maximum area $A=100$ . Thus the dimension should be $l=w=10$ .