A rectangular lot is bordered on one side by a building and the other 3 sides by $800$m of fencing. Determine the area of the largest lot possible.
$\endgroup$1 Answer
$\begingroup$Hint: Let $l$ denote length and $w$ denote width, and $a$ denote area.
$2l+w=800$
You also know that $lw=a$
Find $w$ in terms of $l$ (or $l$ in terms of $w$) and substitute it into the area formula.
So, $w=800-2l$
Sub this in, giving you: $a=l(800-2l)$
Can you go on from here?
If you need more help, mouse over the area below.
$\endgroup$ 2Expand $a=l(800-2l)\implies 800l-2l^2$. Since the coefficent on the $l^2$ is negative that means there will be a maximum, because it will be concave down. Looking at the equation above, you can see that the $l$-intercepts are $400$ and $0$. You can get those $l$-intecepts by setting $l(800-2l)=0$. You can then see that $800-2l=0\implies-2l=-800\implies l=400$. For the other $l$-intercept you can see that it'd be $l=0$. Since a parabola is symmetric, you can do $\frac{400+0}{2}=200$ to get the $l$-coordinate of the vertex. Note that this is important because this is when the maximum occurs: at $l=200$. You then sub $l=200$ into your area formula, giving you: $a=200(800-2(200))=200(800-400)=200(400)=80000$.
