A triangle has a base of 16 inches and an altitude of 8 inches. Find the dimensions of the largest rectangle that can be inscribed in the triangle if the base of the rectangle coincides with the base of the triangle.

Answer 1

I suspect it will be 4x8, but let's see what the calculations show.

If triangle ABC has vertices at (0,0),(16,0) and (h,8) then the sloping sides are lines with the equations

y = 8/h x
y = 8(x-16)/(h-16)

Now, suppose the inscribed rectangle has height k. The the line y=k intersects the sides where x is

hk/8 and k(h-16)/8 + 16

The length of the rectangle is thus 16-2k

The area of the rectangle is

a = k(16-2k) = 16k-2k^2
The vertex of this parabola is where k=4.

Thus the inscribed rectangle of maximal area is half the height of the triangle, making its base half the base of the triangle. The rectangle is thus 4x8.