Largest rectangle that can fit isnide of equilateral triangle

Question

I have an equilateral triangle with a fixed side-length $x$. What is the largest area of a rectangle that can be put inside the triangle?

Answer 1

Consider the situation as in the image..

Then the side $b$ depends on $a$ by the formula $b = \tan(60°) \frac{x-a}{2} = \sqrt{3}\frac{x-a}{2}$, so the area of the rectangle is $S = ab = a\sqrt{3}\frac{x-a}{2}$.

Now look for the maximum of $S$ as a function of $a$ (calculate the first derivative and find the zero point or notice it is a parabola, so its maximum is in the middle of zero-points). In both cases you get that the maximum is in the point $a = \frac{x}{2}$, hence the maximum area is $\frac{x}{2}\sqrt{3}\frac{x-\frac{x}{2}}{2}=\frac{x^2\sqrt{3}}{8}$.

Note that the area of the entire triangle $ \frac{x^2\sqrt{3}}{4}$, so the maximum rectangle area is $1/2$ the triangle area.

Answer 2

Let $a$ be the length of the side of the rectangle, which placed on the side of our triangle, and $b$ be the length of another side of the rectangle.

Thus, $$\frac{\frac{x\sqrt3}{2}-b}{\frac{x\sqrt3}{2}}=\frac{a}{x},$$ which gives $$b=\frac{\sqrt3}{2}(x-a).$$ Id est, by AM-GM we obtain: $$S_{rectangle}=ab=\frac{\sqrt{3}}{2}a(x-a)\leq\frac{\sqrt3}{2}\left(\frac{a+x-a}{2}\right)^2=\frac{x^2\sqrt3}{8}.$$ The equality occurs for $$a=x-a,$$ which says that we got a maximal value.