Question

A piece of wire of length L will be cut into two pieces, one piece to form a square and the other piece to form an equilateral triangle. How should the wire be cut so as to: (a) Maximize the sum of the areas of the square and the triangle? (b) Minimize the sum of the areas of the square and the triangle?

Answer 1

Answer:

  (a)  square: L; triangle: 0.

  (b)  square: L·(-16+12√3)/11; triangle: L·(27-12√3)/11

Step-by-step explanation:

Strategy: First we will write each area in terms of its perimeter. Then we will find the total area in terms of the amount devoted to the square. Differentiating will give a way to find the minimum total area.

__

In terms of its perimeter p, the area of a square is ...

  A_square = p^2/16

In terms of its perimeter p, the area of an equilateral triangle is ...

  A_triangle = p^2/(12√3)

Then the total area of the two figures whose total perimeter is L with "x" devoted to the square is ...

  A_total = x^2/16 + (L-x)^2/(12√3)

__

(a) We know when polygons are regular, the one with the most area for the least perimeter is the one with the most sides. Hence, the total area is maximized when all of the wire is devoted to the square.

__

(b) The derivative of A_total with respect to x is ...

  dA/dx = x/8 -(L-x)/(6√3)

This will be zero when ...

  x/8 = (L-x)/(6√3)

  x(6√3) = 8L -8x

  x(8 +6√3) = 8L

  x = L·8/(6√3 +8) = 8L(6√3 -8)/(64-108)

  x = L·(12√3 -16)/11

The total area is minimized when L·(12√3 -16)/11 is devoted to the square, and the balance is devoted to the triangle.