The maximum possible area of a rectangle given a perimeter of 16 units can be found by recognizing that, for a fixed perimeter, a square gives the maximum area.

Step-by-step solution:

1. The formula for the perimeter of a rectangle is: $P = 2 \times (L + W)$ where $L$ is the length and $W$ is the width. Given $P = 16$, we have: $2 \times (L + W) = 16$ Simplifying this: $L + W = 8$

2. The area of the rectangle is: $A = L \times W$ To maximize the area, we express $W$ in terms of $L$ using $L + W = 8$: $W = 8 - L$ So, the area becomes: $A = L \times (8 - L)$ Expanding this: $A = 8L - L^2$

3. To find the maximum area, take the derivative of the area function with respect to $L$ and set it to zero: $\frac{dA}{dL} = 8 - 2L = 0$ Solving for $L$: $2L = 8 \implies L = 4$

4. Since $L + W = 8$, if $L = 4$, then $W = 4$. Therefore, the rectangle is a square with side lengths of 4 units.

5. The area of this square is: $A = 4 \times 4 = 16 \text{ square units}$

Thus, the maximum possible area of a rectangle with a perimeter of 16 units is 16 square units. None of the answer choices listed are correct.

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Related questions:

1. What is the relationship between perimeter and area for different rectangles?
2. How does maximizing the area of a rectangle relate to optimization in calculus?
3. How do you calculate the perimeter of a square?
4. How does the area of a rectangle change if one side is doubled?
5. How would the maximum area change if the perimeter were increased?

Tip: For a fixed perimeter, the rectangle with the largest area is always a square.