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JonF
Apr6-04, 11:59 PM
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...

arildno
Apr7-04, 03:22 AM
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?

TheMadCapBeta
Apr7-04, 01:07 PM
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.

Integral
Apr7-04, 05:23 PM
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