A rectangular field is to be enclosed on four sides with a fence. Fencing costs $8 per foot for two

opposite sides, and $2 per foot for the other sides. Find the dimensions of the field of area 900 ft2
that would be the cheapest to enclose.

square root of 225 is 15

Let's assume the length of the field is L ft and the width is W ft.

The area of the field is given as 900 ft², so we have the equation:

L * W = 900

To find the dimensions that would be the cheapest to enclose, we need to minimize the cost of the fencing.

The cost of fencing the two opposite sides, which have length L, is given as $8 per foot. So the cost for these two sides would be 2 * 8 * L = 16L.

The cost of fencing the other two sides, which have length W, is given as $2 per foot. So the cost for these two sides would be 2 * 2 * W = 4W.

The total cost of the fencing is the sum of the costs for the opposite sides and the other sides:

Total cost = 16L + 4W

To minimize the cost, we can express one of the variables in terms of the other using the equation for the area, and substitute it into the total cost equation.

Let's solve the area equation for one of the variables:

L * W = 900

W = 900 / L

Now we can substitute W into the total cost equation:

Total cost = 16L + 4(900 / L)

To find the dimensions that would minimize the cost, we need to find the values of L that minimize the total cost. We can do this by finding the derivative of the total cost equation and setting it equal to zero.

Taking the derivative of the total cost equation with respect to L:

d(Total cost)/dL = 16 - 3600 / L²

Setting it equal to zero and solving for L:

16 - 3600 / L² = 0

Simplifying:

16L² - 3600 = 0

Dividing by 16:

L² - 225 = 0

Factoring:

(L - 15)(L + 15) = 0

Setting each factor equal to zero and solving for L:

L - 15 = 0 or L + 15 = 0

L = 15 or L = -15

Since the length cannot be negative, we discard L = -15.

Therefore, the length of the field is L = 15 ft.

Now we can substitute this value back into the equation for the width:

W = 900 / L
W = 900 / 15
W = 60 ft

So the dimensions of the field that would be the cheapest to enclose are length = 15 ft and width = 60 ft.

To find the dimensions of the field that would be the cheapest to enclose, we can start by considering the costs for the different sides of the fence.

Let's assume the length of the field is L ft and the width of the field is W ft.

The cost of the two opposite sides (length) would be $8 per foot, so the total cost for those sides would be 2L * $8 = $16L.

The cost of the other two sides (width) would be $2 per foot, so the total cost for those sides would be 2W * $2 = $4W.

The total cost to enclose the field would then be the sum of the costs for the two opposite sides and the costs for the other two sides:

Total Cost = $16L + $4W

We are given that the area of the field is 900 ft^2, so we have the equation:

L * W = 900

To find the dimensions of the field that would be the cheapest to enclose, we need to minimize the total cost function with respect to L and W.

To do this, we can substitute the value of W from the area equation into the total cost equation:

Total Cost = $16L + $4(900/L)

Now we can take the derivative of the total cost function with respect to L and set it equal to zero to find the critical points:

d(Total Cost) / dL = 16 - 3600/L^2

Setting this equal to zero and solving for L, we get:

16 - 3600/L^2 = 0
L^2 = 3600/16
L^2 = 225
L = ±√225
L = ±15

We discard the negative value since lengths cannot be negative, so L = 15 ft.

Substituting this value back into the area equation, we can determine the corresponding width:

15 * W = 900
W = 900 / 15
W = 60 ft

Therefore, the dimensions of the field that would be the cheapest to enclose are 15 ft by 60 ft.

C = 2x + 2y

C = 16x + 4y
A = xy
900 = xy
y = 900/x

C = 16x + 4(900/x)
C = 16x + 3600/x
C' = 16 - 3600/x^2
0 = 16 - 3600/x^2
16 = 3600/x^2
16x^2 = 3600
x^2 = 225
x = 25
y = 900/15
y = 60