assuming a rectangular fenced area, let
parallel length = x
area = x(1000-x)/3 = 1/3 (1000x-x^2)
Just figure where the vertex of the parabola is, and that will give you the maximum area.
Old McDonald has 1000 feet of fencing, and wants to build a pen along the side of his barn that will have two separate areas- one for his pigs, and one for his cows. if only one of the fences is to run parallel to the barn, find the dimensions of the pen that would maximize the area.
4 answers
where did you get the 1/3 from
Assume the length of the fence opposite the barn side is: x.
The remaining (1000-x) feet of fence work is divided into three by the two other sides and the partition. Each being: (1000-x)/3.
The area of the pen is: y = x(1000-x)/3
The area is maximum when the derivative is zero. The derivative is:
y' = (1000-2x)/3
Solve for x when y' = 0.
The remaining (1000-x) feet of fence work is divided into three by the two other sides and the partition. Each being: (1000-x)/3.
The area of the pen is: y = x(1000-x)/3
The area is maximum when the derivative is zero. The derivative is:
y' = (1000-2x)/3
Solve for x when y' = 0.
since this was pre-calc, I assumed that calculus was not available. But the parabola's vertex is from back in Algebra I!