# Maximal area of equilateral triangle inside rectangle.

If the perimeter of the rectangle is P, what would be the maximal area of the equilateral triangle if: - One of the sides of the triangle coincides with one of the sides of the rectangle - We remove this condition and the equilateral triangle is merely inscribed in the rectangle.

1st part: The dimensions can be written as $$x$$ and $$p-x$$. For this, the equilateral triangle would need to have side length equal to the minimum of $$x$$ and $$p-x$$ so it can fit in. If I assume that the minimum is $$x$$, then the area of the triangle is just $$\frac{\sqrt{3} x^2}{4}$$ that does not have a minimum. I am not sure how to maximise the function if $$p-x$$ is the smaller dimension.

2nd part: Mathworld says that it needs to be inclined at $$15 ^\circ$$ and doesn’t say why this is, or how they calculated the maximal area.

Any help appreciated! Thanks

• is the triangle inscribed inside the rectangle with one of its sides coinciding with one of rectangle's?Does inside the rectangle means it perfectly fits inside I.e. is the vertex touching the opposite side? – ADITYA PRAKASH Mar 19 at 6:42
• Yes, sorry, I edited the typo. I just want the triangle with maximal area, whether or not each of its vertices touches one side of the rectangle. – Andrew Mar 19 at 7:05
• I think you are referring to this page: mathworld.wolfram.com/EquilateralTriangle.html – David K Mar 19 at 12:54
• The MathWorld page does not say the triangle must be inclined at 15 degrees within the rectangle. The 15 degrees is to fit the largest equilateral triangle inside a square. – David K Mar 19 at 12:58

That's how you can do the first part. For the second part, instead of putting one side of the triangle exactly on one side of the rectangle, you can try putting an angle $$\theta$$ between them. For a given $$\theta,$$ make the rectangle's perimeter as small as possible. What is the resulting figure? What is the rectangle's perimeter as a function of $$\theta$$? How do you minimize it?
You can also do the problem in the more straightforward way you suggest, where $$p = \frac12P$$ (a fixed value) and the sides of the rectangle are $$x$$ and $$p - x.$$ You must still have a clear picture of what you're doing. Since the choice of which side to label $$x$$ is arbitrary, I would suggest starting the first part by first assuming $$x$$ is the length of the side of the rectangle that lies along one side of the triangle, and then try to figure out whether this is the larger or smaller side of the rectangle. (Better still, just solve for $$x,$$ and that will tell you.)