Calculus Question - Find dimensions

Question

The sum of the perimeters of an equilateral triangle and a square is 20. Find the dimensions of the triangle and the square that produce a minimum total area.

Answer 1

There are several ways to set up this problem.

Let x = the perimeter of the equilateral triangle.

Then 20 - x = the perimeter of the square.

The length of one side of the triangle is x/3. The length of one side of the square is (20 - x)/4.

The formulas for area:

Equilateral triangle: A = s²√3/4. Square: A = s². (s represents the side length of each figure).

A_triangle = (x/3)²√3/4 = x²√3/36

A_square = [(20 - x)/4]² = (1/16) (20 - x)²

TotalArea = A = (√3/36)x² + (1/16)(20 - x)²

A'(x) = (√3/18)x + (1/8)(20 - x)(-1)

Set A' equal to zero:

(√3/18)x = (1/8)(20 - x)

Multiply both sides by 72:

4 √3 x = 9(20 - x)

4√3 x + 9x = 180

x(4√3 + 9) = 180

x = 180/(4√3 + 9) ≈ 11.3

Since x is the perimeter of the triangle, the side length of the triangle is: 60/(4√3 + 9).

The perimeter of the square is 20 - [180/(4√3 + 9)] = 80√3/(4√3 + 9).

The side length of the square is: 20√3/(4√3 + 9)

Check:

3(60/(4√3 + 9) + 4(20√3/(4√3 + 9) = [(180 + 80√3) /(4√3 +9)] = 20[(4√3 + 9)/(4√3 + 9)] = 20

Note that for completeness it should be shown that the critical number found by setting A' = 0 gives a minimum value for the area (perhaps by showing that A'(1) is negative (A is decreasing) and A'(12) is positive (A is increasing). So a minimum at x ≈ 11.3.