We have 900 ft of fencing and we want to construct a back yard. If we are using the building as part of the barrier for the yard what are the dimensions that allow maximum area?
I am not even sure how to start though I know I am supposed to calculate the derivative of the setup. Thank you
if the yard has dimensions x and y, and side x is parallel to the building, then the fencing used is
x+2y = 900
The area is
a = xy = (900-2y)*y = 900y - 2y^2
da/dy = 900 - 4y
da/dy=0 when y=900/4 = 225
so, x = 900-2y = 450
the yard is thus 450 by 225
I do not think I made it clear sir. The part where I was stuck was where 100 ft. of the building is covered by the side of the yard.
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| yard |
| |
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} building {
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the building length is 100 feet
Sorry for any confusion this caused.
As the text drawing did not come out clear I hope this may clarify what I meant.
imgur /F5NJVqp
You're right - you did not make it clear that the building was only 100 feet wide.
If the building on;y covers 100 feet of the side x, then
x+(x-100)+2y = 900
2x+2y=1000
x = 500-y
and follow the same steps above.
If I still have it wrong, then just draw yourself a diagram and label the various parts, making sure all the fence sections add up to 900. Then substitute for x or y in the formula a=xy.
To find the dimensions that allow for the maximum area, you need to maximize the area function with respect to the dimensions of the yard.
Let's assume the back yard is rectangular, with one of its sides being the side of the building. Let's call the length of the back yard L (in feet) and the width of the back yard W (in feet).
The perimeter of the back yard is the sum of the lengths of all four sides, which includes the building side. Since we have 900 ft of fencing, we can write the equation:
L + 2W = 900 (Equation 1)
The area of a rectangle is given by multiplying its length and width, so the area of the back yard is:
A = LW (Equation 2)
To solve for the dimensions that maximize the area, we need to eliminate one of the variables in Equations 1 and 2. We can do this by solving Equation 1 for L in terms of W:
L = 900 - 2W (Equation 3)
Substituting Equation 3 into Equation 2, we get:
A = (900 - 2W)W
Now, we have a function for the area in terms of a single variable W. To find the dimensions that maximize the area, we can take the derivative of the area function with respect to W, set it equal to zero, and solve for W.
dA/dW = 900 - 4W = 0
Now, solve for W:
900 - 4W = 0
4W = 900
W = 225
Therefore, W = 225 ft.
Substituting this value of W back into Equation 3, we can find the corresponding length L:
L = 900 - 2W
L = 900 - 2(225)
L = 900 - 450
L = 450
Therefore, L = 450 ft.
So, the dimensions that allow for the maximum area are a width of 225 ft and a length of 450 ft.