# Thread: Maximise a rectangle inside a triangle

1. ## Maximise a rectangle inside a triangle

An isosceles triangle of width and height 4m has a rectangle of height $\displaystyle h$ and width $\displaystyle w$ inside it such that the top corners are always flush to the sides of the triangle. What is the maximum area for the rectangle.

I know the area of the triangle is $\displaystyle 8$ and have a function for the area of the rectangle in terms of $\displaystyle w$ and $\displaystyle h$

Gives me $\displaystyle A(w,h) = 8 - f(w,h)$

The only help in need is another relationship between w and h so I can have the area in terms of one variable either $\displaystyle w$ or $\displaystyle h$. Then I can make $\displaystyle A'=0$ , solve that and I will be finished.

2. Originally Posted by Bushy
An isosceles triangle of width and height 4m has a rectangle of height $\displaystyle h$ and width $\displaystyle w$ inside it such that the top corners are always flush to the sides of the triangle. What is the maximum area for the rectangle.

I know the area of the triangle is $\displaystyle 8$ and have a function for the area of the rectangle in terms of $\displaystyle w$ and $\displaystyle h$

Gives me $\displaystyle A(w,h) = 8 - f(w,h)$

The only help in need is another relationship between w and h so I can have the area in terms of one variable either $\displaystyle w$ or $\displaystyle h$. Then I can make $\displaystyle A'=0$ , solve that and I will be finished.
Take the bottom-left right-angled triangle or the bottom-right one.

It's base is $\displaystyle \frac{4-w}{2}$ and it's height is $\displaystyle h$

$\displaystyle tan\theta=\frac{4}{2}=2=\frac{h}{\left(\frac{4-w}{2}\right)}=\frac{2h}{4-w}$

3. This is using the ratios of the sides of 2 similar triangles?

Therefore $\displaystyle \frac{4}{2} = \frac{h}{2-\frac{1}{2}w}\implies 4-w=h$

4. Yes,

which is the same as $\displaystyle tan(corner\ angle)$