Answer:
s = 3: perimeter 12, area 9
Step-by-step explanation:
We are interested in side lengths s such that ...
P ≥ A
4s ≥ s²
4s -s² ≥ 0
s(4 -s) ≥ 0
This is the equation of a parabola that opens downward. I has zeros at s=0 and s=4, so will have a value at least 0 for 0 ≤ s ≤ 4.
A square of side length 3 will have a perimeter greater than the area.
