Posted by **keeter** on Monday, January 17, 2011 at 8:50pm.

a rectangle is to be inscribed in a isosceles triangle of height 8 and base 10. Find the greatest area of such rectangle.

- alg 2 -
**Reiny**, Monday, January 17, 2011 at 9:43pm
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 = 5-x

let PE = y

by similar triangles:

y/(5-x) = 8/5

y = (40-8x)/5

Area of rectangle = 2xy

= 2x(40-8x)/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

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