A farmer has 100m of wire mesh to fence a rectangular enclosure. What is the greatest area he can enclose with the wire mesh?

Let the length of the rectangular enclosure be $L$ meters and the width be $W$ meters.
The total length of wire mesh is the perimeter of the rectangle.

$$2L + 2W = 100$$

The area of the rectangular enclosure is given by:

$$A = L \times W$$

From the perimeter equation, we can simplify:

$$L + W = 50$$

Express $L$ in terms of $W$:

$$L = 50 - W$$

Substitute this into the area equation:

$$A = (50 - W)W$$ $$A = 50W - W^2$$

To find the maximum area, we differentiate the area function $A$ with respect to $W$ and set the derivative to zero.

$$\frac{dA}{dW} = \frac{d}{dW}(50W - W^2)$$ $$\frac{dA}{dW} = 50 - 2W$$

Set $\frac{dA}{dW} = 0$:

$$50 - 2W = 0$$ $$2W = 50$$ $$W = 25 \text{ m}$$

To confirm this is a maximum, we can check the second derivative:

$$\frac{d^2A}{dW^2} = -2$$

Since the second derivative is negative, the value $W=25$ m corresponds to a maximum area.

Substitute $W = 25$ m back into the equation for $L$:

$$L = 50 - W = 50 - 25 = 25 \text{ m}$$

Now, calculate the maximum area:

$$A_{max} = L \times W = 25 \text{ m} \times 25 \text{ m}$$ $$A_{max} = 625 \text{ m}^2$$

The greatest area the farmer can enclose is $\text{625 m}^2$.