b) Work out the value of $$ x $$ that means that the rectangle has the maximum area. If your answer is a decimal, give it to 1 d.p. c) Explain what you notice about the side lengths of this rectangle.

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To solve the problem, we need to find the value of $$ x $$ that maximizes the area of a rectangle. Let's assume the rectangle has a length $$ l $$ and a width $$ w $$, and that the area $$ A $$ is given by the formula:

$$
A = l \cdot w
$$

### Step 1: Define the relationship between the dimensions

If we have a fixed perimeter $$ P $$ for the rectangle, we can express one dimension in terms of the other. For example, if the perimeter is given by:

$$
P = 2l + 2w
$$

We can rearrange this to express $$ w $$ in terms of $$ l $$:

$$
w = \frac{P}{2} - l
$$

### Step 2: Substitute into the area formula

Now, substituting $$ w $$ into the area formula gives:

$$
A = l \left( \frac{P}{2} - l \right)
$$

This simplifies to:

$$
A = \frac{P}{2}l - l^2
$$

### Step 3: Find the maximum area

To find the maximum area, we can take the derivative of $$ A $$ with respect to $$ l $$ and set it to zero:

$$
\frac{dA}{dl} = \frac{P}{2} - 2l
$$

Setting the derivative equal to zero:

$$
\frac{P}{2} - 2l = 0
$$

Solving for $$ l $$:

$$
2l = \frac{P}{2} \implies l = \frac{P}{4}
$$

### Step 4: Find $$ w $$

Now, substituting $$ l $$ back into the equation for $$ w $$:

$$
w = \frac{P}{2} - l = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}
$$

### Conclusion

Thus, both the length and width of the rectangle that maximizes the area are equal:

$$
l = w = \frac{P}{4}
$$

This means that the rectangle is actually a square when the area is maximized.

### Answer to part b

The value of $$ x $$ that maximizes the area is $$ \frac{P}{4} $$. If $$ P $$ is known, we can calculate $$ x $$ and provide it to one decimal place.

### Answer to part c

The observation about the side lengths of this rectangle is that they are equal when the area is maximized, indicating that the rectangle is a square. This is a general property of rectangles: for a given perimeter, the square has the maximum area.
To maximize the area of a rectangle with a fixed perimeter, both the length and width should be equal, making the rectangle a square. Therefore, the value of $$ x $$ that maximizes the area is $$ \frac{P}{4} $$.

b) Work out the value of \( x \) that means that the rectangle has the maximum area. If your answer is a decimal, give it to 1 d.p. c) Explain what you notice about the side lengths of this rectangle.

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