A court is to be surrounded by plastic stripping 30m long. One side of the court does not need the stripping because it is a brick wall. What must be the dimensions of the court in order to obtain a maximum area? What is the maximum area? Thank you! :)

To find the dimensions of the court that result in a maximum area, we need to apply the concept of optimization. Let's assume the court has a length of L meters and a width of W meters.

We know that the total length of the plastic stripping is 30 meters, and one side of the court (the length) does not require the stripping since it is a brick wall. Therefore, the total length of the three sides that require the stripping is 30 - L meters.

The perimeter of the court is the sum of the lengths of all four sides, which is given by: P = 30 - L + 2W.

Since the perimeter is constant at 30 - L meters, solving this equation in terms of one variable will give us the dimensions that maximize the area.

The formula for the area (A) of a rectangle is given by A = L * W.

Now, let's express the width (W) in terms of L and P:
P = 30 - L + 2W
2W = P - 30 + L
W = (P - 30 + L) / 2

Substituting this value of W into the area formula:
A = L * ((P - 30 + L) / 2)

Expanding and simplifying the equation:
A = (PL - 30L + L^2) / 2

To find the maximum area, we need to differentiate A with respect to L and set it equal to zero:
dA/dL = (P - 30 + 2L) / 2 = 0

Solving for L:
P - 30 + 2L = 0
2L = 30 - P
L = (30 - P) / 2

Now we can substitute this value of L back into the equation for the area to find the maximum area (A):
A = ((30 - P)/2)(P - 30 + (30 - P)/2)

Simplifying further:
A = ((30 - P)/2)(30 - P)

To find the maximum area, we need to find the critical points of A. Let's take the derivative of A with respect to P:
dA/dP = (P - 30)/2 - (30 - P)/2

Setting this equal to zero and solving for P:
(P - 30)/2 - (30 - P)/2 = 0
P - 30 - 30 + P = 0
2P - 60 = 0
2P = 60
P = 30

So, the critical point is P = 30, which means the maximum area occurs when the perimeter is 30 meters.

Substituting P = 30 into our equation for L:
L = (30 - 30)/2
L = 0/2
L = 0

We have L = 0, which indicates that the length of the court would be zero in order to maximize the area. However, this doesn't make sense practically. It suggests that the maximum area is obtained when the court is just a point.

Thus, it seems there might be a mistake or missing information in the problem statement, as the given information doesn't lead to a valid solution for a rectangular court maximizing the area.

The maximum area would be a square, 3 sides of 10m.