Question

A farmer wants to fence an area of 6 millon square feet in a rectangutor field and then divide it in half with a fence paraliel to one of the sides of the rectangle. Let y represent the length (in feet) of a side perpendicular to the dividing fence, and let x represent the length (in feet) of a side parallel to the dividing fence. Let Frepresent the fengeh of fencing in feet. Write an equation that represents F in terms of the vanable x. F(x)= Find the derluative F′(x), f−1(x)= Find the critical numbers of the function, (Enter your answers as a coerma-separated list. If an answer does not exist, enter owE) What should the lengttis of the sides of the rectangular field be ( in ft) in order to minimize the cost of tho fence? smaller vahue targer value.

Answer 1

• F(x) = 2x +18×10⁶/x
• F'(x) = 2 -18×10⁶/x²
• {-3000, 0, 3000} — critical numbers
• 2000 ft, 3000 ft — dimensions

Explanation:

You want an equation for the length of fence, its derivative, its critical numbers, and the values of the field dimensions for a rectangular area of 6M square feet that is x feet long and y feet wide with a dividing fence that is also y-feet long. The dimensions minimize the fence length.

Fence length

The field is rectangular with one side being x and the other being y. The area is ...

Area = x·y = 6×10⁶

The total length of fence is ...

F(x, y) = 2x +3y

Using the area relation we can write y in terms of x:

y = 6×10⁶/x

So, the length of fence required is given by the function ...

F(x) = 2x + 3(6×10⁶/x)

F(x) = 2x + 18×10⁶/x

Derivative

The derivative can be found using the power rule. It is ...

F'(x) = 2 -18×10⁶/x²

Critical numbers

The critical numbers are the values of x that make the derivative undefined or zero. With x in the denominator, the derivative will be undefined when x=0.

Solving F'(x) = 0, we have ...

0 = 2 - 18×10⁶/x²

18×10⁶ = 2x² . . . . . multiply by x², add 18×10⁶

9×10⁶ = x² . . . . . . divide by 2

±3000 = x . . . . . . square root

The critical numbers are {-3000, 0, 3000}.

Dimensions

The length of fence is minimized when x = 3000 and y = 6×10⁶/3000 = 2000.

The field is 2000 ft by 3000 ft.

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Additional comment

You will note that the total lengths of fence in the x- and y-directions are the same. They are both 6000 feet, half the total length of fence required. This is the generic solution to this sort of cost minimizing problem.

For this single-partition case, the long side is x = √(3A/2) for area A. In general, the x dimension is √(Ny/Nx·A) where Nx and Ny are the numbers of fence segments in each direction. This remains true if one side has no fence segment because a "river" or "barn" is used as the barrier.

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