1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $$2f(2)=32$$

$\endgroup$
  • $\begingroup$ Thank you for showing me how to solve it! You helped me understand it. Enjoy your green check! $\endgroup$ – Dylan Apr 25 '14 at 0:27
2
$\begingroup$

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 $0\le x\le \sqrt{12}$.

$\endgroup$
0
$\begingroup$

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 $h$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.