# A farmer wants to enclose a 450,000m^2 rectangular field with fence. She then wants to sub-divide this field into three smaller fields by placing additional lengths of fence parallel to one of the side. How can she do this so that she minimizes the cost?

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My equations are 450,000=xy and P=6x+4y, where x is the width and y is the length. However my teachers equation was different for the perimeter, his was P=4x+2y

My equations are 450,000=xy and P=6x+4y, where x is the width and y is the length. However my teachers equation was different for the perimeter, his was P=4x+2y

The minimum perimeter is given when

Without loss of generality let us assume that the field is divided as shown:

The area enclosed is given as

# xy=450000 => y=450000/x#

The total perimeter of all fencing is given by:

As indicated wlog we could interchange

# P=4x + 2*450000/x #

# \ \ \ =4x + 900000/x #

Differentiating wrt

# (dP)/dx =4 - 900000/x^2 #

At a critical point we have

# => 4 - 900000/x^2 = 0 #

# :. 900000/x^2 = 4 #

# :. x^2 = 900000/4 #

# :. x^2 = 225000 #

# :. x = 150sqrt(10) ~~ 474.3 => y ~~ 948.7#

This gives us

We can visually verify that this corresponds to a minimum by looking at the graph:

graph{4x + 900000/x [-10, 1000, -1000, 7830]}

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To minimize the cost of enclosing the field and subdividing it into three smaller fields, the farmer should divide the rectangular field into two equal-sized smaller rectangles by placing a fence parallel to one of the sides. This creates two rectangles each with an area of 225,000m^2. Then, the farmer should place two additional fences parallel to the same side, dividing one of the smaller rectangles into two equal-sized squares, each with an area of 112,500m^2. This results in three smaller fields, each with an area of 112,500m^2.

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To minimize the cost, the farmer should divide the rectangular field into three equal smaller fields. This can be achieved by placing two additional lengths of fence parallel to the shorter side of the rectangle. This will create three smaller rectangles of equal size, each with an area of 150,000m^2. This configuration minimizes the amount of fencing needed to enclose the entire area while creating three equal-sized fields.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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