# What is the maximum possible area of the rectangle that is to be inscribed in a semicircle of radius 8?

Aug 29, 2015

$64$

#### Explanation:

Firstly, draw the rectangle in the semicircle such that its center lies on the center of the diameter of the circle. Draw two lines from that center to the point where the rectangle intersects the arc of the semicircle.

Denote this length as $8$, the radius of the semicircle. Denote the angle between them as $\theta$

Then the area of the triangle bounded by the radii and the rectangle is given by $\Delta = \frac{1}{2} \left(8\right) \left(8\right) \sin \theta = 32 \sin \theta$

The are of the rectangle is twice the angle of the triangle, $A = 64 \sin \theta$

$\frac{\mathrm{dA}}{d \theta} = 64 \cos \theta$

At maximum, $\frac{\mathrm{dA}}{d \theta} = 0$, thus $\cos \theta = 0 , \theta = \frac{\pi}{2}$ since within the semicircle, $0 < \theta < \pi$

Therefore $A = 64 \sin \left(\frac{\pi}{2}\right) = 64$