Algebra 2
posted by Sarah .
Find the dimensions and area of the largest rectangle that can be inscribed in a right triangle whose sides are 9cm, 12cm, 15cm.
Use quadratic functions to solve.

Put the points of the triangle at (0,0), (0,9) and (12,0)
Let the insribed rectangle have corners at:
(0,0), (x,0), (x,y'), and (0,y')
y' is on the hypotenuse of the triangle, so that
y = 12  (4/3)x
So y' = 12  (4/3)x
The rectangle area is
A(x) = x*y' = 12 x  (4/3)x^2
There is a maximum when dA/dx = 0
12 = (8/3) x
x = (3/8)*12 = 4.5
y' = 12  (4/3)(9/2) = 5
The maximum rectangle area is xy'= 22 and the side lengths are 4.5 and 5. 
This is normally a challenging calculus problem, however it can be worked with similar triangles, or algebra.
First, note the triangle is a right triangle. Lay it out so the 12 dimension is on the x axis, and the 9 is on the y axis.
Note the equation of the hypotenuse is
y=9/12 x + 9
let one side of the rectangle going upward as h, and the side along the x axis as W. So the intersection with the hypotenuse is w,9w/12 + 9
So the drill is to find the max area.
h= 9/12 w + 9
Area= hw= 9/12 w^2+9w
so when is Area max? 
I calculated the point on the hypotenuse incorrectly. It is
y' = 12  (4/3)(9/2) = 6
The rectangle side lengths are 4.5 and 6, and the area is 27. That equals half the area of the triangle, (1/2)*9*12 = 54. BobPursley tells me that Euclid proved this in many different ways, without calculus.
Thanks to Bob for pointing this out 
A package with square ends has a combined length and girth (girth is the perimeter of a cross section) of 120 in. The surface area of the entire package is 3600 sq. in.
Determine the dimensions of thr package. s in.*s in.*l in.
one solution is: 11.08in*11.08in*75.68in.
find the other solution
hint: if 4S + length=120, then the length=1204S
Round your answer to 2 decimals
Enter the 3 dimensions separated by comas