alg 2
posted by keeter on .
a rectangle is to be inscribed in a isosceles triangle of height 8 and base 10. Find the greatest area of such rectangle.

Did you make a sketch?
Draw triangle ABC so that AB = AC
Draw the height AD, D on BC
Let P be the point where the rectangle touches AC
Draw PE perpendicular to BC, E on BC
let DE = x, then EC = 5x
let PE = y
by similar triangles:
y/(5x) = 8/5
y = (408x)/5
Area of rectangle = 2xy
= 2x(408x)/5
= (80x  16x^2)/5
d(Area)/dx = (80  32x)/5 = 0 for max of area
32x = 80
x = 80/32 = 5/2
y = (40  8(2.5))/5 = 4
greatest area = 2xy = 2(5/2)(4) = 20