# Algebra 2

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Find the dimensions and area of the largest rectangle that can be inscribed in a right triangle whose sides are 9cm, 12cm, 15cm.

• Algebra 2 -

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.

• Algebra 2 -

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?

• Algebra 2 (correction) -

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

• Algebra 2 -

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=120-4S