Finding Max Area of Rectangle in Isos. Triangle

In summary, the problem asks to find the maximum possible area of a rectangle that can be placed inside an isosceles triangle with base 6 and height 12. The solution involves finding a linear relationship between the base and height of the rectangle, which is y = -2x + 12, and using calculus to find the maximum area of the rectangle, which is 3 by 6.
  • #1
Kyber
3
0

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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  • #2
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:
  • #3
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?
 
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  • #4
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
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
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
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.
 

What is an isosceles triangle?

An isosceles triangle is a triangle that has two sides of equal length. This means that two of its angles are also equal in measure.

What is the maximum area of a rectangle inscribed in an isosceles triangle?

The maximum area of a rectangle inscribed in an isosceles triangle is when the rectangle's base is equal to the base of the triangle and its height is half the length of the triangle's other two sides.

How do you find the maximum area of a rectangle in an isosceles triangle?

To find the maximum area of a rectangle in an isosceles triangle, you can use the formula A = (b/2)(h/2), where A is the area, b is the base of the triangle, and h is the height of the rectangle.

Why is the maximum area of a rectangle in an isosceles triangle important?

The maximum area of a rectangle in an isosceles triangle is important because it allows us to find the largest possible rectangle that can fit inside the triangle. This can be useful in various applications, such as maximizing storage space or optimizing design layouts.

Can the maximum area of a rectangle in an isosceles triangle be greater than the area of the triangle itself?

No, the maximum area of a rectangle in an isosceles triangle cannot be greater than the area of the triangle itself. This is because the rectangle is inscribed within the triangle, meaning it cannot have a larger area than the space it is contained in.

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