A rectangular enclosure is to be constructed with 2,000 feet of fence in order to have four sections, as in the diagram below. What should be the dimensions in order to maximize the area enclosed? What is the maximum area? (So the diagram is just four rectangles connected side by side.
Sort of like | | | | | but with tops and bottoms too).
I totally drew a blank on this I'm sorry for not showing my work I don't know where to start.
well, if each pen is x by y, then the vertical lines have length x, and the horizontal lines have length 4y, then we know
5x+8y = 2000
then the area is
a = 4xy = 4x(2000-5x)/8 = 1000x - 5/2 x^2
This just a parabola, and its vertex will be the maximum area. Work it out, and you will see that as in all problems like this, the area is maximum when the fencing is divided equally among lengths and widths.
In this case, that means 1000 ft for the 5 vertical fences (x=200) and 1000 ft for the horizontal fences (y=125).
So the maximum area is the vertex? Is the vertex 200, 1000?
And the dimensions in order to maximize the area enclosed are 200, 125?
well of course the vertex is the maximum area. The parabola represents the area, right?
And you are correct in your final answers, with the exception that when x=200 and y=125, 4xy = 4*200*125 = 100,000
No problem! I can help you out with this problem. To find the dimensions that would maximize the area enclosed by the rectangular enclosure, we can use optimization techniques.
Let's start by assigning variables to the dimensions of the enclosure. Let the width of each section be 'x' and the height be 'y'. Since there are four sections, the total width of the enclosure would be 4x.
To calculate the perimeter of the enclosure, we need to consider both the vertical and horizontal sides. The vertical sides contribute to the perimeter as 2y, and the horizontal sides contribute as 4x. Therefore, our equation for the perimeter is:
2y + 4x = 2000 (equation 1)
Now, because we want to maximize the area, we need to formulate an equation for the area of the enclosure. The area can be calculated as the product of the width and the height, which is (4x)(y) = 4xy.
To find the maximum area, we need to eliminate one of the variables in the area equation. We can use equation 1 to solve for y and substitute it back into the area equation. Let's rearrange equation 1 to solve for y:
2y = 2000 - 4x
y = (2000 - 4x)/2
y = 1000 - 2x (equation 2)
Now substitute equation 2 into the area equation:
Area = 4x(1000 - 2x)
Area = 4000x - 8x^2
Now we have a quadratic equation for the area of the enclosure. To maximize the area, we can take the derivative of the area equation with respect to x and set it equal to zero. Let's find the derivative:
d(Area)/dx = 4000 - 16x
Setting the derivative equal to zero:
4000 - 16x = 0
16x = 4000
x = 250
Now we have the value of x that maximizes the area. To find the corresponding value of y, we can plug x = 250 into equation 2:
y = 1000 - 2(250)
y = 500
So, the dimensions that maximize the area enclosed are a width of 250 feet and a height of 500 feet for each section. The total area can be calculated by multiplying the width (4x) by the height (y):
Maximum Area = 4x * y
Maximum Area = 4 * 250 * 500
Maximum Area = 500,000 square feet
Therefore, the maximum area enclosed by the rectangular enclosure is 500,000 square feet.