# How to find the dimensions of a rectangle if its area is to be a maximum?

A rectangle has two of its vertices as the $x$-axis. The other two vertices are on the parabola whose equation is $y=18-x^2$.

What are the dimensions of the rectangle if its area is to be a maximum?

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Amira: It is desirable that you do not simply put the homework tag, as it does not provide sufficient information for potential answerers to have any idea of the level of your mathematical background. Is this algebra homework? Precalculus? Calculus? Please add another tag to give us the relevant information. –  Cameron Buie Jun 10 '12 at 19:08
I see Marvis has added the Calculus tag for you. If this is accurate, then my answer should be sufficient for your purposes (comment if you are still uncertain what to do). If it is not Calculus homework, then you should remove that tag and place the appropriate one, and comment on my answer to let me know you need more help. –  Cameron Buie Jun 10 '12 at 19:14

The equation for the area of the rectangle (depending on the $x$-coordinate of the vertex on the right side of the $y$-axis) will be $A(x)=2x(18-x^2)$. Do you know how to determine which $x$ maximizes this? Note that we will only be considering $0<x<\sqrt{18}$.
Note that the height of the rectangle is $y$, and the width is $x-(-x)=2x$, so the area is $2xy$. We also know $y=18-x^2$, since $(x,y)$ lies on that parabola, so that gives us the expression I have above for area, and it's entirely in terms of $x$. We can also rewrite it as $A(x)=36x-2x^3$. Taking the derivative, we have $A'(x)=36-6x^2$. We need to find the positive solution to $A'(x)=0$, which will give us the maximum area. If $x_0$ is that value of $x$, then the dimensions will then be $2x_0$ by $18-x_0^2$.
Think of Cameron's $A(x)$ as just some function in $x$. Then in order to obtain local maximum or local minimum, one would take its $\_\_\_\_\_$. –  math-visitor Jun 10 '12 at 19:14