Let's break down and solve the problem step by step.

Problem Summary:

You have a wire that is 24 cm long, and it is cut into two pieces. Each piece is bent to form a square. The side length of one square is $x$ cm. You are asked to:

• (a) Find a function that gives the total area of the two squares in terms of $x$.
• (b) Find the side length $x$ that minimizes the total area of the two squares.
• (c) Find the minimum area of the two squares.

Solution:

(a) Finding the total area function $A(x)$

Let:

• $x$ be the side length of the first square.
• $y$ be the side length of the second square.

The wire has a total length of 24 cm, so the total perimeter of both squares must equal 24 cm. The perimeter of a square is 4 times the side length.

Thus, the perimeter constraint is: $4x + 4y = 24$ Dividing the equation by 4: $x + y = 6 \quad \text{or} \quad y = 6 - x$

The areas of the squares are:

• Area of the first square: $A_1 = x^2$
• Area of the second square: $A_2 = y^2 = (6 - x)^2$

Thus, the total area function $A(x)$ is: $A(x) = x^2 + (6 - x)^2$

Expanding this: $A(x) = x^2 + (36 - 12x + x^2) = 2x^2 - 12x + 36$

So, the total area function is: $A(x) = 2x^2 - 12x + 36$

(b) Minimizing the total area

To find the value of $x$ that minimizes the total area, we take the derivative of $A(x)$ and set it equal to zero.

The derivative of $A(x)$ is: $A'(x) = 4x - 12$

Set $A'(x) = 0$ to find the critical points: $4x - 12 = 0$ $x = 3$

Thus, the side length $x = 3$ cm minimizes the total area.

(c) Minimum area

Substitute $x = 3$ into the total area function $A(x)$: $A(3) = 2(3)^2 - 12(3) + 36$ $A(3) = 2(9) - 36 + 36 = 18$

So, the minimum total area is $18$ square centimeters.

Final Answers:

• (a) The total area function is $A(x) = 2x^2 - 12x + 36$.
• (b) The side length $x = 3$ cm minimizes the total area.
• (c) The minimum area enclosed by the two squares is 18 square centimeters.

Would you like more details or have any questions about this solution? Here are 5 related questions you could explore:

1. How do you verify if a critical point gives a minimum or maximum value?
2. What happens if the wire's total length changes to a different value, like 30 cm?
3. How would you solve this problem using a graphical method?
4. Can the total area function $A(x)$ ever be negative?
5. What if the wire was divided into more than two pieces—how would the solution change?

Tip: When working with quadratic functions like $A(x) = 2x^2 - 12x + 36$, knowing the shape of the parabola (whether it opens up or down) can help you quickly identify if you're dealing with a minimum or maximum value.