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Mathematics, optimization exercise : maximize the area of a field with a given perimeter. You will have to find the dimensions of the field (width and length, in meters) that has a maximum area with a perimeter of 1200 m. How to do so ? Maximizing the quadratic equation Area, using first and second derivative.
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1.3 Optimization: A farmer wants to set up a fence around his field

A farmer whishes to build a fence around his rectangular field and to divide it in two equal parts as well. The dividing fence must be parallel to one of the side of the field. You are asked to find the dimensions of the field of maximum area that you can fence and divide in two equal parts with a wire fence of 1200 m.

(1) *Area = x . y (in m ^{2})*

*Fence length = 2y + 2x + x = 1200 m*

*1200 = 2y + 3x*

*y = (1200 - 3x)/2*

(2) *y = 600 - 1,5x*

To find the width *x* of the field that makes the area maximum, you must calculate the derivative of the function *Area*, makes it zero and solve for *x*: *Area ' = (x . y) ' = 0*.

*Area = x . y = x . (600 - 1,5x)* by substituting equation (2) for *y*

*Area = 600x - 1,5x ^{2}*

*(Area) ' = (600x - 1,5x ^{2}) ' = 0*

* = 600 - 3x = 0*

*600 = 3x*

*x = 200 m*

We have to check that the x-value *x = 200* corresponds to an extremum that is a maximum and not a minimum because we are trying to maximize the area of the field.
However, in our case the second derivative does not have to be computed because the function *Area = 600x - 1,5x ^{2}* is a quadratic equation (second degree equation), thus it is the equation of a parabola.
Moreover the factor of