# A rectangle with base on the x-axis has its upper vertices on the curve y=12-x^2 . Find the maximum area of such a rectangle.

Okay, so I understand that the equation is a downwards facing parabola with a y-intercept at 12. I don't understand what it means by upper vertices? I know that the answer is 32 but I don't understand how to get there. Can someone please guide me and explain to me the process of solving this problem? Thank you!

Let's $t_1$ and $t_2$ the abscissas of the lower vertices $(t_2>t_1)$ and clearly the upper vertices have the ordinates $$12-t_1^2=12-t_2^2\iff t_1=-t_2\quad\text{since}\; t_1\ne t_2$$ and the area of the rectangle is $$(t_2-t_1) (12-t_1^2)=2t_2(12-t_2^2)$$ hence to answer the question we should maximize the function $$f(t)=t(12-t^2)$$ and since $$f'(t)=12-t^2-2t^2=12-3t^2=0\iff t=\pm2$$ hence we see easily that $t_2=2$ and the area is $$2f(2)=32$$

• Thank you for showing me how to solve it! You helped me understand it. Enjoy your green check! – Dylan Apr 25 '14 at 0:27

Hint: Draw a picture of the downward-facing parabola, and of a rectangle of the type described.

Let $(x,y)$ be the upper right-hand corner of the rectangle. Then by symmetry the base of the rectangle has length $2x$, and the height is $y$, that is, $12-x^2$.

So the area $A(x)$ of the rectangle is given by $$A(x)=2x(12-x^2).$$ Maximize, using the usual tools. Note that we must have $0\le x\le \sqrt{12}$.

The rectangle has $4$ vertices, now the lower two are on the $x$-axis, and the upper two are on a parallel line to the $x$-axis, say at height $h$.

Then, find the $x$ coordinates of the vertices, using $h$: the upper vertices are on the parabole on height $y=h$, so the $x$ coordinates satisfy $h=12-x^2$, i.e. $$x^2\ =\ 12-h$$ Then, you get two solutions, $x_1$ and $x_2$, finally maximize the area of the rectangle ($h\cdot(x_2-x_1)$) in $h$.