Maximum area of a rectangle inside a triangle

[madheretic]

I have to prove that the max area of a rectangle inside a triangle is 1/2 the area of the triangle. I did something that makes me feel like it's wrong, because this is a calc homework and I haven't used any calculus here.

This is what I did.

Area of the rectangle here = 2y (b - 2x).

I have to prove that 2y (b -2x) = (1/2) bh.

So from there, y = (bh) / 4(b - 2x).

So, substituting y in the area of the rectangle, I get 2 [(bh) / 4(2 - bx)] (b - 2x).

By simplifying, I get (bh)/2, thereby proving that the area of this rectangle is 1/2 that of the area of the triangle.

 

[galactus]

I have to prove that the max area of a rectangle inside a triangle is 1/2 the area of the triangle. I did something that makes me feel like it's wrong, because this is a calc homework and I haven't used any calculus here.

This is what I did.

Area of the rectangle here = 2y (b - 2x).

From your diagram, shouldn't that be y(b-2x)?.

I have to prove that y(b -2x) = (1/2) bh.

So from there, y = (bh) / 4(b - 2x).

So, substituting y in the area of the rectangle, I get 2 [(bh) / 4(2 - bx)] (b - 2x). This doesn't equal bh/2

By simplifying, I get (bh)/2, thereby proving that the area of this rectangle is 1/2 that of the area of the triangle.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\(\displaystyle \L\\A_{R}=y(b-2x)\)......[1]

\(\displaystyle \L\\A_{T}=\frac{bh}{2}\)

Use similar triangles:

\(\displaystyle \L\\\frac{b-2x}{h-y}=\frac{b}{h}\)

Solve for y:

\(\displaystyle \L\\y=\frac{2hx}{b}\)

Sub into [1]:

\(\displaystyle \L\\A=\frac{2hx}{b}(b-2x)\)

Differentiate:

\(\displaystyle \L\\\frac{dA}{dx}=\frac{2(bh-4xh)}{b}\)

Set to 0 and solve for x:

\(\displaystyle \L\\\frac{2(bh-4xh)}{b}=0\)

\(\displaystyle \L\\2h=\frac{8xh}{b}\)

\(\displaystyle \L\\x=\frac{b}{4}\)

\(\displaystyle \L\\y=\frac{h}{2}\)

Check: \(\displaystyle \L\\A=(\frac{h}{2})(b-2(\frac{b}{4}))=\frac{bh}{4}\)

Since the area of the rectangle is half the area of the triangle:

\(\displaystyle \L\\2\frac{bh}{4}=\frac{bh}{2}\)....The area of the triangle.

 

[madheretic]

You're right, I mixed up the values in between. I was looking at a wrong group of calculations instead of the diagram when I typed them.

Thanks, I get the idea now.