What is the largest possible area of the field? You need to use calculus to find your solution.

Question

A farmer has 600 ft of fencing to enclose a rectangular field which he plans to subdivide into two pens of equal area, with a single length of fencing running down the middle. What is the largest possible area of the field? You will need to use calculus to find your solution.

I got the constraint equation, 3x+2y=600 and the optimization equation, A=xy. Should I differentiate the constraint equation?

Answer 1

You want to write the optimization equation in terms of one variable. So solve the constraint equation for y (or x) and substitute into A = xy.

For example:

2y = 600 - 3x

y = 300 - (3/2)x

So,

A = x[300 - (3/2)x]

A(x) = (-3/2)x² + 300x

The graph of that function is a parabola (downward opening), so you do not really need calculus to solve the problem. Locate the vertex with whatever method, like the axis of symmetry is given by x = -b/2a or in this case: x = -300/-3 = 100.

You arrive at the same critical number by finding A'(x) and setting the derivative equal to zero:

A'(x) = -3x + 300

-3x + 300 = 0

x = 100

Apply 1st (or 2nd) derivative test to show that critical number generates a maximum area.

Since y = 300 - (3/2)(100 = 150, the max area is 100(150) = 15000 sq. ft.

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