MHB Maximizing Area of Inscribed Rectangle - Yahoo Answers

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The problem involves maximizing the area of a rectangle inscribed under the curve y=7/(1+x^2) with one side on the x-axis. The area function is derived as A(x)=14x/(x^2+1), where the base is 2x and height is y. By finding the derivative and setting it to zero, the critical point is determined to be x=1. The first derivative test confirms that this point corresponds to a maximum area. The vertices of the rectangle with maximum area are located at (-1,0), (-1,7/2), (1,0), and (1,7/2).
MarkFL
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Here is the question:

Optimization, Maximum area...?

A rectangle has one side on the x-axis and two vertices on the curve

y=7/(1+x^2)

Find the vertices of the rectangle with maximum area.

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

Because of the even symmetry of the given curve, the base $b$ of the rectangle will be $b=2x$ and the height $h$ will be $y$. And so the area of the rectangle is:

$$A=2xy$$ where $0\le x$

Substituting for $y$, we then get the area as a function of $x$:

$$A(x)=2x\left(\frac{7}{x^2+1} \right)=\frac{14x}{x^2+1}$$

Now, we want to find the critical value(s), so we equate the derivative with respect to $x$ to zero:

$$A'(x)=\frac{\left(x^2+1 \right)(14)-(14x)(2x)}{\left(x^2+1 \right)^2}=\frac{14\left(1-x^2 \right)}{\left(x^2+1 \right)^2}=0$$

And so we see that our relevant critical value is:

$$x=1$$

Using the first derivative test, we can see that on $(0,1)$ the area function is increasing and on $(1,\infty)$ the area function is decreasing, so we know our critical value is at a maximum. Hence the vertices of the rectangle of maximal area are:

$$\left(-1,0 \right),\,\left(-1,\frac{7}{2} \right),\,\left(1,0 \right),\,\left(1,\frac{7}{2} \right)$$
 
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