**Question:**

11. A rectangular field is to be fenced on three sides using a total of 50 m of fencing material. The side along the river does not need to be fenced. Determine the dimensions of the field that will maximize the enclosed area.

**Answer:**

To solve this problem, we need to find the dimensions of the rectangular field that will maximize the enclosed area. We are given that the field is to be fenced on three sides using a total of 50 m of fencing material, and the side along the river does not need to be fenced.

**Step 1: Define the variables**

Let's define the variables:
- Length of the field = L
- Width of the field = W

**Step 2: Formulate the constraints**

Since the side along the river does not need to be fenced, the total fencing material required will be the sum of the length and two times the width:
2W + L = 50

**Step 3: Formulate the objective function**

The objective is to maximize the enclosed area, which is given by:
Area = Length × Width = L × W

**Step 4: Solve the system of equations**

To find the dimensions that maximize the enclosed area, we need to solve the system of equations formed by the constraints and the objective function.

From the constraint equation, we can express L in terms of W:
L = 50 - 2W

Substituting this expression for L in the objective function, we have:
Area = (50 - 2W) × W = 50W - 2W^2

**Step 5: Maximize the area**

To maximize the area, we need to find the value of W that maximizes the objective function. We can do this by taking the derivative of the objective function with respect to W, setting it equal to zero, and solving for W.

d(Area)/dW = 50 - 4W

Setting this derivative equal to zero and solving for W, we get:
50 - 4W = 0
4W = 50
W = 12.5 m

**Step 6: Find the dimensions of the field**

Substituting the value of W into the constraint equation, we can find the length of the field:
2W + L = 50
2(12.5) + L = 50
25 + L = 50
L = 25 m

Therefore, the dimensions of the field that will maximize the enclosed area are:
Length = 25 m
Width = 12.5 m