Finding Max Area of Rectangle in Isos. Triangle

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Homework Help Overview

The problem involves finding the maximum area of a rectangle that can fit inside an isosceles triangle with a base of 6 and a height of 12. The context is geometric optimization within the constraints of the triangle's dimensions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the rectangle's dimensions and the triangle's geometry, exploring linear equations to express height in terms of base. Some express confusion about deriving a second equation necessary for optimization.

Discussion Status

Several participants have shared their attempts at formulating equations and using calculus to find the maximum area. There is an ongoing exploration of the relationships between the dimensions of the rectangle and the triangle, with some participants questioning their understanding of the setup and the equations involved.

Contextual Notes

Participants mention the need to account for the entire width of the triangle when deriving relationships, indicating a potential misunderstanding of the triangle's symmetry and dimensions. There is also a reference to the time spent on the problem, suggesting varying levels of confidence among participants.

Kyber
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Homework Statement


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.


Homework Equations


None.


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?
 
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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:
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:
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
 
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.
 
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
 
Kyber said:
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

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

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