# Calculus

a rancher has 1200 fet of fencing to enclose two adjacent cattle pens. What dimensions should be used so that the enclosed area is maximized?

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1. You don't describe the diagram, but I am familiar with this type of question.
Assuming there is a common width between the two fields
let the width be x ft
let the length of the combined field be y
so
3x + 2y = 1200
y = 600 - (3/2)x

area = xy
= x(600 - (3/2)x)
= 600x - (3/2)x^2
d(area)/dx = 600 - 3x
= 0 for a max of area
3x = 600
x = 200
y = 600-(3/2)(200) = 300

state the conclusion

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