# I'm haveing optimization problems

1. Apr 6, 2004

### JonF

This is homework (forgive me) but I don’t want an answer I would just like to know what I am doing wrong.

Here is the problem:

Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3cm and 4cm if tow dies of the rectangle lie along the legs.

Here is what I did

L = length of rectangle
W = width
Theta = angle 4 leg of the right triangle

I’m trying to optimize W*L where :
W < 3
L < 4

These are the equations I got:
Tan(theta) = w/(4-L)
Tan(theta) = (3-W)/L
Tan(theta) = 3/4

W = $3(4-L)/4$

so:

{3 - 3(4-L)/4}/ L = 3/4

and that is as far as this brain will take me...

Last edited: Apr 7, 2004
2. Apr 7, 2004

### arildno

Basically, what you have done "wrong" is back-substituting, so that you end up with the trivially correct equation 3/4=3/4 .
You already have a perfectly good expression for the width, w=3(4-L)/4.
With this expression, what is the area of the rectangle?
How can you find the maximum of this area?

3. Apr 7, 2004

After you figure out that one, try this one.

Basically the same problem, but one side of the rectangle is on the hypotenuse of the triangle.

4. Apr 7, 2004

### Integral

Staff Emeritus
Set up a coordinate system with the right angle at, (4,0) thus the line representing the hypotenuse is given by y= 3x/4.

Let L and H be the sides of your rectangle. One corner must be on the line y(x) = 3x/4
one side will be L=4-x the other H=y

The area is A=L*H = (4-x)*y = (4-x)*3x/4

Compute $$\frac {dA} {dx}= 0$$