MHB Andrew's question at Yahoo Answers regarding maximizing the area of a rectangle

[MarkFL]

Here is the question:

A rectangle has one side on the x-axis and two vertices on the curve y=4/4+x^2?

Find the vertices of the rectangle with maximum area.
Vertices =
Enter your answers as a comma-separated list of ordered (x,y) pairs, e.g., (1,0),(8,0),(1,4),(8,4).

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Andrew,

The base of the rectangle will be $b=2x$ where $0\le x$ and the height of the rectangle will be $h=\dfrac{4}{4+x^2}$. Hence the area function is:

$$A(x)=2x\cdot\frac{4}{4+x^2}=\frac{8x}{x^2+4}$$

To find our critical values, we may differentiate this area function with respect to $x$ and equate the result to zero:

$$A'(x)=\frac{\left(x^2+4\right)(8)-(8x)(2x)}{\left(x^2+4\right)^2}=\frac{8(2+x)(2-x)}{\left(x^2+4\right)^2}=0$$

The non-negative critical value here is:

$$x=2$$

And we see that the derivative is positive to the left of this value and negative to the right, and so by the first derivative test we may conclude that this critical value is at a relative maximum.

Thus, our vertices are:

$$\bbox[5px,border:2px solid #207498]{\left(-2,\frac{1}{2}\right),\,\left(2,\frac{1}{2}\right),\,(2,0),\,(-2,0)}$$