Answer:
- maximum area: 11,250 m²
- dimensions: 150 m parallel to the beach, 75 m out from the beach
Step-by-step explanation:
Let x represent the length of the swimming area along the beach. After x amount of floats are used in that direction, the remaining (300-x) of floats will be split in two to form the ends of the swimming area. The total area is the product of these dimensions:
A = LW
A = x(300 -x)/2
You will recognize this as a factored quadratic with zeros at x=0 and x=300. The vertex of a quadratic is found halfway between the zeros, at ...
x = (0+300)/2 = 150
In this quadratic, A = -1/2x^2 +150x, the leading coefficient is negative, so the graph opens downward and the vertex is a maximum.
The maximum area is ...
A = (150)(300 -150)/2 = 11,250 . . . square meters
The dimension along the beach is 150 m; the dimension out from the beach is (300 -150)/2 = 75 m.
_____
Additional comment
This is an example of a "maximize the area for a given cost" problem that involves a rectangular area. The generic solution to such a problem is that the sum of costs in one direction is equal to the sum of costs in the orthogonal direction.
Here, half the total cost (length of float) is parallel to the beach, and the other half is perpendicular to the beach. If the area had any partitions, their cost would be added to the total for the direction of their layout.
