Arectangle is to be inscribed in a right triangle having sides of length 66in, 88in, and 110in. find the dimensions of the rectangle with greatest area assuming the rectangle is positioned as in the accompanying figure.

 This one is hard to explain in print. Let x be the side of the rectangle on the hypotenuse and y be the other side. There is a triangle formed at the right angle of the original which is similar to the original. 
So x is to 10 as z is to 8: x/10 = z/8 
and z = 4/5 x 
The upper part of that leg would be 8 - 4/5 x 

The triangle at the top is also similar: 
So y is to 6 as (8 - 4/5 x) is to 10: y/6 = (8-4/5 x)/10 
and y = 3/5 (8- 4/5 x) = 24/5 - 12/25 x 

Now area = xy 
Deriv = 0 and solve 

Where is the problems