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

Question

Christopher Allers · Accepted Answer

#64#

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 = 1/2 (8)(8) sin theta =32 sin theta#

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

#(dA)/(d theta) = 64 cos theta#

At maximum, #(dA)/(d theta) = 0#, thus #cos theta = 0, theta = pi/2# since within the semicircle, #0 < theta < pi#

Therefore #A = 64 sin (pi/2) = 64#

Ezekiel Alexander · Answer

undefined To find the maximum area of the rectangle inscribed in a semicircle, the rectangle's width should be equal to the semicircle's diameter, and its height should be half of the semicircle's radius.

Given the radius of the semicircle, $r = 8$, the diameter is $2r = 16$, and half the radius is $r/2 = 4$.

Therefore, the maximum possible area of the rectangle is $16 \times 4 = \boxed{64}$ square units.