Homework Help: Finding Max Area of Rectangle in Isos. Triangle

1. Dec 12, 2007

Kyber

1. The problem statement, all variables and given/known data
An isosceles triangle has base 6 and height 12. Find the maximum possible area of a rectangle that can be placed inside the triangle with one side on the base of the triangle.

2. Relevant equations
None.

3. The attempt at a solution
Well, so far I've created an X-Y Axis and made the triangle symmetric to the Y-Axis. I am, however, really bad at these and need help from here. Where should I go from here?

2. Dec 12, 2007

Kyber

I know:

Area of Rectangle = Max
A = lw

I found through the Pythagorean theorem that 3^2 + 12^2 = 12.37 (The length of the hypotenuse)

I don't know what my secondary equation should be. That's my question.

As for effort, I've spent 45 minutes on this problem trying to figure it out. I don't know how to get my second equation. Please help!

Last edited: Dec 12, 2007
3. Dec 12, 2007

Dick

If the rectangle is inside of the triangle there is a linear relation between it's base and height. For example if it's base is 3 then it's height is 6. If base=0 then height=12, if base=6 then height=0. Do you see why? And can you figure it out for any base?

Last edited: Dec 12, 2007
4. Dec 13, 2007

Kyber

Yes, I totally understand that. I got the answer to be 3 by 6 through guess and check, but I need to prove this through calculus.

How do I get my second equation.

I have A=L*W, but that's all. I need a second equation so I can get to the max through calculus, not guess and check.

The linear equation is -2x+12=y

5. Dec 13, 2007

phagist_

I got the equation as y= -4x+12 for the right hand slope of the triangle (for x>0)

and y is also the height of the rectangle at point x, so getting the A=X*Y in terms of X;
as X=x and Y=-4x+12 ;

A= x*(-4x+12) which becomes A= -4x^2+12x, then differentiating for 0;
x=0 or x=1.5

x cannot = 0 as then the rectangle would have no base width, hence x=1.5.

subbing x=1.5 into our previous equation (y= -4x+12) to get the height, hence the height is 6.

But because we only used one half of the isosceles triangle (i.e y= -4x+12 only governs the right hand slope of the triangle ) we must double the x value to get the FULL width which is 2*1.5 = 3.

6. Dec 13, 2007

Office_Shredder

Staff Emeritus
If x is the base, and y the height, then the relationship of x from the midpoint of the triangle to one of its ends and y is y=-4x+12 x from 0 to 3. But the base is in fact twice as long, so the relationship is in fact y=-2x+12 x from 0 to 6. Though I see you doubled at the end to account for the discrepancy

7. Dec 13, 2007

Dick

Ok then. Substitute the linear equation into the area equation A=L*W=x*y=x*(-2x+12). Now use calculus.