A rectangular play area is to be fenced off in a person's yard and is to contain 108 yd^2. The next-door neighbor agrees to pay half the cost of the fence on the side of the play area that lies along the property line. What dimensions will minimize the cost of the fence?
cost is proportional to perimeter, so we want to minimize the perimeter enclosing 108 yd^2. If x is the length of the fence on the property line, 108/x is the other dimension.
p(x) = 2(x + 108/x)
dp/dx = 2(1 - 108/x^2)
dp/dx=0 when x^2 = 108
x = 6√3 = 10.39
the pool is thus
6√3 by 6√3
as we know, a square has maximum area for a given perimeter; conversely a square has minimum perimeter for a given area.
Now, if we want to minimize the cost to the pool owner, that would involve subtracting half the cost of the side on the property line:
p(x) = 2(x + 108/x) - x/2
dp/dx = 2(1 - 108/x^2) - 1/2
dp/dx=0 when x = 12
now the fence would be 12 by 9
The longer border length makes the neighbor pay more.
Well, if we're talking about minimizing the cost of the fence, then I guess we should go for the smallest possible dimensions.
Now, let's assume that the length of the play area is L yd and the width is W yd. We know that the area is 108 yd^2, so we can write the equation L * W = 108.
Since the next-door neighbor is willing to pay half the cost of the fence on one side, we can assume that the fence will be placed on the longer side, i.e., the length.
To minimize the cost of the fence, we should aim for the smallest possible length. In other words, let's make L as small as possible.
Given that L * W = 108, if we make L as small as possible, then W has to be as large as possible. So, let's make W as large as possible while maintaining the area of 108 yd^2.
Now, there are multiple combinations of L and W that can give us an area of 108 yd^2, but one of the pairs that minimizes the cost of the fence could be L = 12 yd and W = 9 yd.
Why? Well, 12 * 9 = 108, and this combination has the smallest possible length (L) and the largest possible width (W) while maintaining the area of 108 yd^2.
Of course, keep in mind that this is just one possible solution, and there might be other pairs of L and W that also minimize the cost of the fence. But hey, at least we've got one option that fits the bill!
To minimize the cost of the fence, we need to find the dimensions of the play area.
Let's assume the length of the play area is L and the width is W.
We are given that the area of the play area is 108 yd^2, so we can write the equation:
L * W = 108
Next, we need to find the cost of the fence. The fence cost is divided into two parts:
1. The cost for the side of the fence along the property line, which the neighbor will pay for.
2. The cost for the other three sides of the fence, which the person will pay for.
Let's say the cost of fencing per yard is C.
The cost of the fence along the property line (which the neighbor will pay for) is 1/2 of the total length of the play area (L) at C per yard:
(L/2) * C
The cost of the other three sides of the fence is 2 times the length (L) plus 2 times the width (W) at C per yard:
2L + 2W
The total cost of the fence is the sum of the two costs:
Total Cost = (L/2) * C + 2L + 2W
Now, we need to minimize this cost function.
To find the dimensions that will minimize the cost, we can use calculus. We take the derivative of the cost function with respect to L and W, then set them equal to zero and solve for L and W.
First, let's find the derivative of the cost function with respect to L:
d(Cost)/dL = (C/2) - 2
Next, let's find the derivative of the cost function with respect to W:
d(Cost)/dW = 2
Setting both derivatives equal to zero:
(C/2) - 2 = 0 and 2 = 0
Solving the first equation:
(C/2) = 2
C = 4
Now, substituting the value of C in the area equation:
L * W = 108
L * W = 108
Plugging in C = 4 and simplifying:
L * W = 108
L * W = 108
Finding the factors of 108, we can find possible dimensions for L and W:
1 x 108
2 x 54
3 x 36
4 x 27
6 x 18
9 x 12
We can see that the dimensions (L, W) that will minimize the cost of the fence are (9, 12).
To minimize the cost of the fence, we need to find the dimensions of the play area that will maximize its area. Let's assume the length of the play area is "l" yards and the width is "w" yards.
The area of a rectangle is given by the formula A = length × width.
Given that the area is 108 yd^2, we have the equation:
l × w = 108
Now, we are told that the next-door neighbor will pay for half of the cost of the fence on the side of the play area that lies along the property line. This means we want to minimize the cost of the fence on the other three sides. Let's assume the cost of the fence on each of the other three sides is "c" dollars per yard. The total cost of the fence, C, is given by the equation:
C = 2c(2l + w) + c(l)
Simplifying this equation, we have:
C = 4cl + 2cw + cl
C = 5cl + 2cw
To minimize the cost, we need to minimize the function C(l, w). Now we can express C in terms of a single variable, either l or w, to find the minimum value.
Using the equation relating the area and the dimensions, l = 108 / w, we can substitute l in the expression for C:
C = 5(108 / w) + 2cw
To find the minimum, we take the derivative of C with respect to either l or w and set it equal to zero. Let's differentiate C with respect to w:
dC / dw = 0 - 5(108 / w^2) + 2c
Setting this expression equal to zero and solving for w:
- 5(108 / w^2) + 2c = 0
5(108 / w^2) = 2c
w^2 = (5 * 108) / (2c)
w = √[(5 * 108) / (2c)]
Now, we know that l = 108 / w, so substituting the above expression for w:
l = 108 / √[(5 * 108) / (2c)]
These equations give us the dimensions of the play area (l and w) that will minimize the cost of the fence (C).