How should this be done?
Suppose you have 132 m of fencing with which to make two side-by-side rectangular enclosures against an existing wall. if the rectangular enclosures are adjacent and of the same depth what is the maximum are that can be enclosed?
Draw a diagram!
You will need three cross fences, and one that is parallel to the wall.
Let the length of each cross fence be x.
Express the length L of the longitudinal fence in terms of x and 132 m.
Express the total area A(x) of the two rectangular areas in x and L, and subsequently in x only.
Use calculus to maximize A(x). Solve for x.
To solve this problem, we need to find the dimensions of the rectangular enclosures that will maximize the enclosed area.
Let's start by understanding the problem statement. We have 132 meters of fencing, which we will use to create two side-by-side rectangular enclosures against an existing wall. The enclosures are adjacent and have the same depth. We want to determine the maximum area that can be enclosed.
Let's denote the width of each enclosure as "x" meters and the depth as "y" meters. Since the enclosures are adjacent, they will share a common wall. Therefore, the total length of the two enclosures combined would be x + x = 2x meters.
Now, we need to consider the fencing. The total fencing used would be the sum of the lengths of the two enclosures and the shared wall, which is 2x + y meters. According to the problem, this total fencing length is 132 meters.
So, we have the equation: 2x + y = 132.
Next, we need to express the area in terms of x and y. The area of a rectangle is given by the formula: Area = length × width, which in this case is (2x) × (y) = 2xy.
To maximize the area, we need to find the optimum values of x and y that satisfy the constraint equation and maximize 2xy.
To do this, we can use the constraint equation to solve for one variable in terms of the other and substitute it into the area equation. Let's solve the constraint equation for y:
2x + y = 132
y = 132 - 2x
Now, substitute this value of y into the area equation:
Area = 2x(132 - 2x) = 264x - 4x^2.
To find the maximum area, we can take the derivative of the area equation with respect to x, set it equal to zero, and solve for x:
d/dx (264x - 4x^2) = 264 - 8x.
264 - 8x = 0
8x = 264
x = 264/8
x = 33.
Now that we have the value of x, we can substitute it back into the constraint equation to find y:
y = 132 - 2x
y = 132 - 2(33)
y = 132 - 66
y = 66.
So, the maximum area that can be enclosed is 2xy = 2(33)(66) = 4356 square meters.
Therefore, to maximize the area, each enclosure should have dimensions of 33 meters by 66 meters.