We are given that Mr. Honda has 80 meters of fencing to enclose a rectangular field, and we are looking for the maximum area he can enclose. The total length of the fencing is used to enclose both the length and the width of the rectangle, and we're given an equation for the area in terms of the length $x$ of the field.

The total perimeter is 80 meters. The perimeter $P$ of a rectangle is given by:

$P = 2x + 2w$ where $x$ is the length and $w$ is the width. We know the perimeter is 80 meters, so:

$2x + 2w = 80$

Solving for $w$:

$w = 40 - x$

The area $A$ of the rectangle is given by:

$A = x \cdot w = x(40 - x)$

Expanding this equation:

$A(x) = 40x - x^2$

This is a quadratic equation of the form $A(x) = -x^2 + 40x$, which opens downward (because of the negative coefficient of $x^2$). The maximum area occurs at the vertex of the parabola.

For a quadratic equation $ax^2 + bx + c$, the x-coordinate of the vertex is given by:

$x = -\frac{b}{2a}$

Here, $a = -1$ and $b = 40$, so:

$x = -\frac{40}{2(-1)} = \frac{40}{2} = 20$

Thus, the length $x = 20$ meters maximizes the area. To find the maximum area, substitute $x = 20$ back into the equation for the area:

$A(20) = 40(20) - (20)^2 = 800 - 400 = 400 \, \text{square meters}$

Therefore, the maximum area is 400 square meters.

The correct answer choice would be: Maximum is 400 ft² (square meters is the correct unit).

Would you like more details or further clarification?

Here are five related questions:

1. What is the vertex formula for a general quadratic function?
2. How would the equation change if the perimeter were different?
3. Can you describe why the quadratic equation represents a parabola?
4. How would the maximum area change if the fencing material were doubled?
5. What happens to the area if the length exceeds 40 meters?

Tip: The vertex of a parabola always gives the maximum or minimum value of a quadratic function, depending on whether it opens upward or downward.