# 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?

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 $0<x<\sqrt{18}$.

Edit: The first thing we probably want to do for a problem like this is draw a picture. A general rectangle as described will look something like this: 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
• No, I don't. I find this very difficult to understand considering I am only in high school. Could you explain it with a little more detail please? – Amira Jun 10 '12 at 19:36
• Ah, I see you are in trigonometry. That does, indeed make this less simple! What topics are being discussed in this section of your text/lectures? That might give me some ideas how you are expected to solve this. – Cameron Buie Jun 10 '12 at 19:46
• Basically using maximum and minimums in equations – Amira Jun 10 '12 at 20:05
• Have you discussed any methods of finding maxima or minima for given functions? – Cameron Buie Jun 10 '12 at 20:10