Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y= 5-x^2. What are the dimensions of such a rectangle with the greatest possible area?

Answer 1

Consider:
Area(A) of rectangle is bh; b=2x and h=y. The area is constrained by y=x^2. So the four 'corners are at x, -x, y=-(5-x^2) and y=5-x^2. A is 2xy. So A=2x(5-x^2). dA/dx = 10-6x^2. Therefore x=+/-(5/3)^1/2. dA=is either the max or minimum area, and since our diagram shows it not to be a minimum, dA=0 is probably the max. This means y=(5-(5/3)^1/2)^2 or 3.333.
The dimensions are 2x=2(5/3)^1/2 or ~2.6, and y~3.333. Area is ~8.66.
To 'check', obtain A for a different value of x (or y) and see that A decreases.

I hope this helps; tell me if I did it wrong.

Answer 2

It makes no sense to include x or y-values less than 0 since they would give a negative area. So by finding where the y-value is 0, we can define a valid area for x:
\[5-x^2=0, x=\pm\sqrt{5} \rightarrow 0<x<\sqrt{5}\]
The base of the rectangle will always have the length 2x, and the height at point x is given by the function 5-x^2. Therefore the area is defined by the function A as follows:
\[A(x)=2x(5-x^2)=10x-2x^3\]
The largest value for A can be found by finding out where A'=0, and making sure A is a maximum. A' is defined as:
\[\frac{d}{dx}(10x-2x^3) = 10-6x^2\]
This can be refactored as:
\[\frac{d}{dx}6(\sqrt{\frac{10}{6}}-x)(\sqrt{\frac{10}{6}}+x)\]
This shows that A' is 0 when:
\[x=0, x=\pm\sqrt{\frac{10}{6}}\]
Since we've said that x has to be greater than 0, the only valid answer is:
\[x=\sqrt{\frac{10}{6}}\]
If we plug this into A, we get:
\[A(\sqrt{\frac{10}{6}})=10\sqrt{\frac{10}{6}}-2(\sqrt{\frac{10}{6}})^3=\frac{20}{3}\sqrt{\frac{10}{6}}\]