Question

A rectangular field having an area of 2700 m2 is to be enclosed by a fence, and an additional fence is to be used to divide the field down the middle. If the cost of the fence down the middle is $6/m, and the fence along the sides costs $9/m, find the dimensions of the field such that the cost of the fencing will be minimized.

Answer 1

To minimize the cost of the fencing for a rectangular field, we can express the total cost in terms of a single variable. By finding the derivative of the cost equation and solving for the variable, we can find the dimensions that minimize the total cost.

Step-by-step explanation:

To minimize the cost of the fencing, we need to find the dimensions of the rectangular field that minimize the total cost of the fencing. Let's assume the length of the field is x meters and the width is y meters. The area is given as 2700 m2, so we have the equation xy = 2700.

The cost of the fence along the sides is $9/m, so the cost of the sides is 2(x+y) * 9 = 18(x+y) dollars. The cost of the fence down the middle is $6/m, so the cost of the middle fence is 6y dollars.

To minimize the total cost, we can express the total cost in terms of a single variable using the equation: Total Cost = 18(x+y) + 6y. We can solve for y in terms of x using the equation xy = 2700, and substitute it back into the total cost equation to get a single variable equation. Then, we can find the value of x that minimizes the total cost.

By finding the derivative of the total cost equation with respect to x, setting it equal to 0, and solving for x, we can find the value that minimizes the total cost. We repeat the same process to find the value of y that corresponds to the minimum cost.

Answer 2

`length = 51.96\;m, breadth = 51.96\;m`

Step-by-step explanation:

Rectangular field:

`Area = l * b` unit^2

`2700 = l * b`

`b=(2700)/(l)` - Eq(A)

The cost of the fence depends on the side lengths of the rectangular fields (the perimeter) plus an additional fence down the middle

`Total\;\;Perimeter = 2l + 2b + l` units

`Cost = 2(6l) + 2(9b) + (6l)` $

`Cost = 18l + 18b` - Eq(B)

Since we want the minimum cost, we need to either differentiate the above equation OR just plot it.

but we need to have only variable on both sides of the equal (=) sign.

so we can substitute Eq(A) in Eq(B)

`Cost = 18l + 18\left((2700)/(l) \right)`

now just simplify it!

`Cost = 18l + (48600)/(l)`

the figure below states that at L = 51.96 meters the cost is least.

by plotting this equation, we'll be able to clearly see the length at which the cost is least! Anywhere else will increase the cost!

We can use
`l = 51.96` to find
`b` from Eq(A)

`b=(2700)/(l)`

`b=(2700)/(51.96)`

`b= 51.96` meters

The dimensions of the rectangle at which the cost of the fence is minimum is::

`length = 51.96\;m, breadth = 51.96\;m`