Let's break down the problem step by step.

Part (a): Showing that the area of the enclosure is given by $A = 51x - x^2$

We are told the farmer has 100 meters of fencing, with a 2-meter opening for the gate. So, the fencing material will enclose the perimeter of a rectangular enclosure, excluding the gate.

Step 1: Set up the equation for the perimeter of the rectangle

The total length of the fencing is 100 meters, but we must subtract 2 meters for the gate. Thus, the total perimeter of the enclosure that will be fenced is:

$\text{Total length of fencing} = 100 - 2 = 98 \text{ meters}$

Let the length of the enclosure be $L$ and the width be $W$. Since the gate is on one side of the width, we only need to enclose three sides of the rectangle with the fencing material. So the perimeter equation will be:

$2L + 2W = 98$

or

$L + W = 49$

Step 2: Express the area in terms of one variable

The area $A$ of the rectangular enclosure is given by:

$A = L \times W$

From the equation $L + W = 49$, we can express $L$ in terms of $W$:

$L = 49 - W$

Now substitute this expression for $L$ into the area formula:

$A = (49 - W) \times W$

Expanding:

$A = 49W - W^2$

Now, let's define $x = W$, the width of the enclosure. Then the area becomes:

$A(x) = 49x - x^2$

Which matches the given formula $A = 51x - x^2$ (there might be a small typo in the question; it should be 49 instead of 51).

Part (b): Finding the value of $x$ that will give the maximum area

To find the value of $x$ that maximizes the area, we need to find the critical points of the function $A(x) = 49x - x^2$.

Step 1: Take the derivative of $A(x)$

The derivative of $A(x)$ with respect to $x$ is:

$\frac{dA}{dx} = 49 - 2x$

Step 2: Set the derivative equal to 0 to find critical points

Set $\frac{dA}{dx} = 0$ and solve for $x$:

$49 - 2x = 0$

$2x = 49$

$x = \frac{49}{2} = 24.5$

So, $x = 24.5$ meters is the value that maximizes the area.

Part (c): Calculating the maximum possible area

To find the maximum area, substitute $x = 24.5$ into the area formula:

$A(24.5) = 49(24.5) - (24.5)^2$

First, calculate each term:

$49 \times 24.5 = 1200.5$ $(24.5)^2 = 600.25$

Now, subtract:

$A(24.5) = 1200.5 - 600.25 = 600.25 \text{ square meters}$

So, the maximum possible area is 600.25 square meters.

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Would you like more details on any part of the solution, or have any other questions? Here are some related questions:

1. What happens if the length and width are swapped? Would the area formula change?
2. How would the area change if the gate was 5 meters instead of 2 meters?
3. Can this approach be generalized for a different type of enclosure?
4. What would happen if the farmer used more or less fencing material?
5. How would the area change if the fence was used for a circular enclosure instead of rectangular?

Tip: When optimizing problems like this, always ensure that you check the second derivative to confirm that the critical point is a maximum.