Final answer:
To minimize the sum of the areas of two squares made from a 10m wire, one must use calculus to derive a function that represents the areas, differentiate it, and solve for the length that yields the minimum sum of areas.
Step-by-step explanation:
The question asks for an optimization strategy for dividing a 10m wire into two parts, each to be bent into a square, such that the sum of the areas of the two squares is minimized. The key here is to use calculus to find the minimum value of a function that represents the sum of the areas.
Let's denote the length of one piece of wire as x meters. Then the other piece will be 10 - x meters long. Since the perimeter of a square is equal to the length of the wire used to create it, the side of the first square will be x/4 meters, and the side of the second square will be (10 - x)/4 meters.
The area of the first square will be (x/4)^2, and the area of the second square will be ((10 - x)/4)^2. Therefore, the total area A is given by:
A = (x/4)^2 + ((10 - x)/4)^2
To minimize the total area, we need to take the derivative of A with respect to x, set it to zero and solve for x. After finding the critical points, we can use the second derivative test or the first derivative test to make sure that the critical point corresponds to a minimum.