1 2 3 4 5 6 7 8. 9 10 Question 10. A farmer has \( \mathbf{2 0 0} \) rnetres of fencing with which to enclose three sides of a rectangular paddock, the fourth side of which is bounded by a river. If the dimensions of the paddock are as shown, what is the value of \( \boldsymbol{x} \) such that the enclosed area is a maximum?

To find the value of \( x \) that maximizes the area of the rectangular paddock, we can set up the problem like this: The farmer has \( 200 \) meters of fencing to enclose three sides of a rectangle. Let the length parallel to the river be \( x \) and the two widths perpendicular to the river be \( y \). From the fencing constraint, we have: \[ x + 2y = 200 \] From this, we can express \( y \) in terms of \( x \): \[ y = \frac{200 - x}{2} \] The area \( A \) of the rectangle is given by: \[ A = x \cdot y = x \cdot \left(\frac{200 - x}{2}\right) = \frac{200x - x^2}{2} \] To maximize the area, we take the derivative of \( A \) with respect to \( x \), set it to zero, and solve for \( x \): \[ \frac{dA}{dx} = \frac{200 - 2x}{2} = 0 \] Solving this gives: \[ 200 - 2x = 0 \implies 2x = 200 \implies x = 100 \] Now, substitute \( x = 100 \) back to find \( y \): \[ y = \frac{200 - 100}{2} = 50 \] So, the dimensions that maximize the area are \( x = 100 \) meters and \( y = 50 \) meters, yielding a maximum area of: \[ A = 100 \cdot 50 = 5000 \text{ square meters} \] In conclusion, the value of \( x \) that maximizes the enclosed area is \( \mathbf{100} \) meters.