Problem solving with derivatives.
Question: A piece of wire length of 30cm is cut into 2 sections. Each section is then bent into the shape of a square. Find the smallest possible value of the sum of the areas of the two squares.
To find the smallest possible value of the sum of the areas of the two squares, we can use the concept of optimization using derivatives. Let's break down the problem into smaller steps:
Step 1: Assign variables
Let's assign variables to the lengths of the sides of the two squares. Let the side length of the first square be x cm and the side length of the second square be (30 - x) cm.
Step 2: Find the areas of the squares
The area of a square is given by the formula A = side length^2. Therefore, the area of the first square is A1 = x^2 cm^2, and the area of the second square is A2 = (30 - x)^2 cm^2.
Step 3: Find the sum of the areas
The sum of the areas is given by S = A1 + A2 = x^2 + (30 - x)^2.
Step 4: Determine the smallest possible value
To find the smallest possible value of the sum of the areas, we need to find the minimum point of the function S(x) = x^2 + (30 - x)^2.
Step 5: Use derivatives to find the minimum
To find the minimum point, we need to find the derivative of the function S(x) with respect to x and set it equal to zero. Then, solve for x to find the critical point(s).
The derivative of S(x) is given by:
S'(x) = d/dx (x^2 + (30 - x)^2)
= 2x - 2(30 - x)
Setting S'(x) equal to zero:
2x - 2(30 - x) = 0
Simplifying the equation:
2x - 60 + 2x = 0
4x - 60 = 0
4x = 60
x = 15
Step 6: Determine the minimum value
To determine that this is the minimum value, we can utilize the second derivative test. If the second derivative is positive, the critical point corresponds to a minimum value.
The second derivative of S(x) is given by:
S''(x) = d^2/dx^2 (2x - 2(30 - x))
= d/dx (2 - 2)
= 0
Since the second derivative is zero, we need to consider higher-order derivatives or perform additional testing. However, in this case, we don't need to proceed further because there is only one critical point.
Step 7: Calculate the areas
Now that we have the critical point, we can substitute it back into the expression for the areas to find the minimum sum of the areas.
For x = 15:
A1 = x^2 = 15^2 = 225 cm^2
A2 = (30 - x)^2 = (30 - 15)^2 = 225 cm^2
Therefore, the smallest possible value of the sum of the areas of the two squares is 225 cm^2 + 225 cm^2 = 450 cm^2.