Maximizing the area of a rectangle formed by bending a wire of length 18 cm into a rectangle, with one side given as x.

To solve the problem of maximizing the area of a rectangle formed by bending a wire of length 18 cm, where one side is given as \( x \), we can proceed as follows:

### Step 1: Express the Perimeter of the Rectangle
The perimeter of the rectangle is given by the length of the wire. Let the length of the other side of the rectangle be \( y \).

\[
\text{Perimeter} = 2x + 2y = 18
\]

### Step 2: Solve for \( y \)
From the perimeter equation:

\[
2x + 2y = 18 \implies x + y = 9 \implies y = 9 - x
\]

### Step 3: Express the Area of the Rectangle
The area \( A \) of the rectangle can be expressed as:

\[
A = x \times y = x \times (9 - x) = 9x - x^2
\]

### Step 4: Maximize the Area
To maximize the area, we need to find the value of \( x \) that maximizes the function \( A(x) = 9x - x^2 \).

We can do this by taking the derivative of \( A(x) \) with respect to \( x \) and setting it equal to zero:

\[
\frac{dA}{dx} = 9 - 2x
\]

Set the derivative equal to zero to find the critical points:

\[
9 - 2x = 0 \implies x = \frac{9}{2} = 4.5 \text{ cm}
\]

### Step 5: Verify that the Critical Point is a Maximum
To verify that \( x = 4.5 \) cm gives a maximum, we can check the second derivative:

\[
\frac{d^2A}{dx^2} = -2
\]

Since the second derivative is negative, the function \( A(x) = 9x - x^2 \) is concave down, meaning \( x = 4.5 \) cm indeed gives a maximum.

### Step 6: Calculate the Maximum Area
Substitute \( x = 4.5 \) cm back into the expression for the area:

\[
y = 9 - 4.5 = 4.5 \text{ cm}
\]
\[
\text{Maximum Area} = 4.5 \times 4.5 = 20.25 \text{ square centimeters}
\]

### Final Answer:
The maximum area of the rectangle is \( 20.25 \) square centimeters, which occurs when both sides of the rectangle are \( 4.5 \) cm, forming a square.

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Would you like more details or have any questions about this problem?

Here are 8 related questions for further exploration:

1. What is the maximum area if the wire length is different, say 24 cm?
2. How would the solution change if the wire forms a triangle instead of a rectangle?
3. What happens to the maximum area if one side is fixed and can't vary?
4. Can the method used above be applied to maximizing the volume of a 3D shape formed by a given wire length?
5. What if the rectangle is not confined to integer dimensions—how does this affect the solution?
6. What is the effect of increasing or decreasing the wire length on the maximum area?
7. How would you solve this problem using the method of completing the square?
8. Can this problem be solved using the method of Lagrange multipliers?

**Tip:** When solving optimization problems, always check the second derivative to confirm whether a critical point is a maximum or minimum.

Learn how to maximize the area of a rectangle formed from an 18 cm wire, with one side fixed at \( x \). Detailed step-by-step solution provided. Suitable for Grades 11-12.

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