Optimization?

Question

A wire of length L is cut into two pieces, one being bent to form a square and the other to form an equilateral triangle. How should the wire be cut if the sum of the two areas is a minimum? How should the wire be cut if the sum of the areas is a maximum?

Answer 1

Let x be the length of the portion of the wire used for the square, and y be the length of the portion used for the triangle.

Then x+y=L. Also, the area of the square is (x/4)^2 = x^2/16.

To get the area of the triangle, note that each side has length y/3. The area is therefore (1/2)(y/3)h, where h is the height. To get the height, use the fact that the triangle can be divided into two right triangles, which are 30-60-90 triangles with hypotenuse y/3 and sides h and y/6. By the pythagorean theorem, h^2 + (y/6)^2 = (y/3)^2, so h^2 = y^2/9 - y^2/36 = y^2/12, or h = y/sqrt(12) = y*sqrt(3)/6.

Therefore the area of the triangle is (1/2)(y/3)(y*sqrt(3))/6 = y^2*sqrt(3)/36.

x+y=L, so the sum of the two areas is

A = x^2/16 + sqrt(3)(L-x)^2/36.

To find the maximum or minimum, we must find critical values of this function. We also have to consider endpoints: if all of the wire is used for the square, then x=L. If all of it is used for the triangle, then x=0.

To find critical values, differentiate, and get

A' = x/8 - sqrt(3)(L-x)/18 = (1/8+sqrt(3)/18)x - sqrt(3)L/18.

Therefore, the only critical value is when x = sqrt(3)L/18 / (1/8+sqrt(3)/18) = sqrt(3)L / (9/4+sqrt(3)).

A'' > 0 for all x, so this value represents a local minimum. Plugging in this critical values, as well as the endpoints x=0 and x=L, and comparing the values of A, will tell you how to cut the wire to maximize or minimize the sum of the areas.