A farmer wants to enclose 2 adjacent rectangular regions next to a river. No fencing will be used next to the river. 60 meters of fencing will be used. What is the area of the largest region that can be enclosed.
I need all of the steps to get to the answer.
See my responses to your other two posts.
To determine the maximum area that can be enclosed, we'll need to apply some mathematical reasoning. Let's break down the problem into steps:
Step 1: Setup the problem
Let's denote the length of one rectangular region as "x" and the length of the other rectangular region as "y". The width of both regions will be the same and can be represented as "w". Since the regions are adjacent, they share a common side of length "w".
Step 2: Create equations
To solve the problem, we need to create equations based on the information provided. The perimeter of the combined rectangular regions is given as 60 meters. Since there are two regions, the total length of fencing used is twice the sum of the lengths and widths. So we can write the equation as:
2(x + y + 2w) = 60
Step 3: Simplify the equation
Simplify the equation by dividing both sides by 2:
x + y + 2w = 30
Step 4: Relate the variables
Since we want to maximize the area, we'll need to express the area in terms of a single variable. We can use substitution to relate the variables. Since one side of each region is shared, we can express the area, A, as:
A = (x + y)w.
Step 5: Solve for one variable
We can solve for one variable by isolating it in terms of the other variable(s). Let's solve for y:
y = 30 - x - 2w
Step 6: Substitute into the area equation
Substituting the value of y from Step 5 into the area equation, we have:
A = [x + (30 - x - 2w)]w
= (30 - 2w)w
Step 7: Maximize the area
To find the maximum area, we need to find the critical points of the area function. This can be done by taking the derivative of the area function with respect to w and setting it equal to zero.
dA/dw = 30 - 4w = 0
Solving the equation, we find:
30 - 4w = 0
4w = 30
w = 7.5
Step 8: Find the corresponding values of x and y
Using the value of w, we can substitute it back into the equation from Step 5 to find the values of x and y:
y = 30 - x - 2w
y = 30 - x - 2(7.5)
y = 30 - x - 15
y = 15 - x
Step 9: Determine the area
Now, we can substitute the values of x, y, and w into the area equation to find the maximum area, A:
A = (30 - 2w)w
= (30 - 2(7.5))(7.5)
= (30 - 15)(7.5)
= 15(7.5)
= 112.5
Therefore, the maximum area that can be enclosed is 112.5 square meters.