The maximum possible area of a yard in terms of $x$

Question

A total of $x$ feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of $x$?

(A) $\frac{x^2}{9}$

(B) $\frac{x^2}{8}$

(C) $\frac{x^2}{4}$

(D) $x^2$

(E) $2x^2$

I do not know the meaning of level rectangular in mathematics, Also I do not know how to divide the problem into smaller ones so that I can solve it in $2.5$ minutes. Could anyone help me please?

Answer 1

You have to form three sides of the rectangle. If the two parallel sides you fence are $a$ the third is $x-2a$, so the area is $a(x-2a)$. Differentiate this with respect to $a$, set to zero, find $a$ as a fraction of $x$, substitute in, and you are done.

Answer 2

Suppose that the side of your rectangle whose opposite side is not bordered by the fence has a length of $s$. Then the two adjacent sides have length $\frac{x-s}{2}$. The area of the rectangle can then calculated to be $s\cdot\frac{x-s}{2}=\frac{sx-s^2}{2}$, so we must maximize $sx-s^2$. With some manipulation we obtain $$sx-s^2=-s^2+2s\cdot \frac{x}{2}-\frac{x^2}{4}+\frac{x^2}{4}=\frac{x^2}{4}-(s-\frac{x}{2})^2$$Now since $x$ is fixed, $\frac{x^2}{4}$ is fixed, too. Hence, we must only minimize $(s-\frac{x}{2})^2$ in order to maximize the area of the rectangle. Since a square is always nonnegative, $(s-\frac{x}{2})^2$ is minimal when it's equal to $0$ which implies $s=\frac{x}{2}$.

Therefore the maximum area of your rectangle is equal to $$\frac{x}{2}\cdot\frac{x-\frac{x}{2}}{2}=\frac{x^2}{8}$$ which we get by chucking $s=\frac{x}{2}$ into the formula for the area of the rectangle we derived above. Note that this solution does not make use of calculus at all.