# Thread: Area of an inscribed Rectangle

1. ## Area of an inscribed Rectangle

A rectangle that is x feet wide is inscribed in a circle of radius 8 ft.

a) Express th area of the rectangle as a function of x.
b) find the domain of the function
c) What dimmensions maximize the area of the rectangle?

I would appreciate help.
thank you

2. Originally Posted by ludiaz223
A triangle that is x feet wide is inscribed in a circle of radius 8 ft.

a) Express th area of the rectangle as a function of x.
b) find the domain of the function
c) What dimmensions maximize the area of the rectangle?

I would appreciate help.
thank you
Always draw a picture

The Point P is located at $(x,\sqrt{64-x^2})$

The red box is one quarter of the area of the rectangle so

$A(x)=4\cdot x \cdot \sqrt{64-x^2}$

The domain is all values of the radicand(the stuff under the square root sign) is greater than or equal to zero.

$64-x^2 \ge 0 \iff (8-x)(8+x) \ge 0$ when we test our values we get $x \in [-8,8]$

To maximize the are we take the derivative of A(x)

$\frac{dA}{dx}=4[\sqrt{64-x^2}+x\cdot \frac{1}{2}\cdot \frac{-2x}{\sqrt{64-x^2}}]=4\left[ \frac{64-x^2-x^2}{\sqrt{64-x^2}}\right]$

setting the numerator equal to zero and solving we get

$64-2x^2=0 \iff x^2=32 \iff x = \pm \sqrt{32}$

Since we are talking about a distance we use only the postive root.

Now when $x=\sqrt{32}$ plugging this into $\sqrt{64-x^2}$ to get the height of the box we get

$\sqrt{64-(\sqrt{32})^2}=\sqrt{64-32}=sqrt{32}$
So the base and the height of the rectangle are the same so it is a square.