Question

The sum of the perimeters of an equilateral triangle and a square is 10. Find the dimensions of the triangle and the square that produce a minimum total area.

Answer 1

Let x be the side of the triangle, y the side of the square. Sum of the perimeters is 10:
3x+4y =10
so
y = (10-3x)/4

Area of square is y^2, area of an equilateral triangle is here: http://www.mathwords.com/a/area_equilateral_triangle.htm

Total of the areas:

\[A = y ^{2} +\frac{ x ^{2}\sqrt{3} }{ 4 }\]

\[A = (\frac{ 10-3x }{ 4 })^{2} +\frac{ x ^{2}\sqrt{3} }{ 4 }\]

Which we differentiate to find where the area is minimum.

Answer 2

\[A = \frac{ (10-3x)^{2} }{ 4^{2} } +\frac{ x ^{2}\sqrt{3} }{ 4 }\]
\[A' = \frac{ 2 }{16 }(-3)(10-3x)+\frac{ 2x \sqrt{3} }{ 4 }\]
\[0 = \frac{ -6 }{16 }(10-3x)+\frac{ x \sqrt{3} }{ 2 }\]
rearranging to find x gives
\[x = \frac{ 30 }{ 9+4\sqrt{3} } = 1.883\]
which you substitute back into y = (10-3x)/4 to find y. x is the length of the triangle's side, y is the length of the square.