# 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

##### 1 Answer

The minimum perimeter is given when

#### Explanation:

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]}