B
billbowden
Guest
"Phil Hobbs" <pcdhobbs@gmail.com> wrote in message
news:2b430f0d-60ca-48eb-bfd7-00edf3b6cbdb@googlegroups.com...
The only calculus I remember is using the first derivative to find the rate
of change. An example is figuring out the optimum dimensions for a pot field
located on a river bank where you havea limited length of fence, say 300
feet. So, if the parallel part of the fence along the river is the x
component, and the perpenducular parts of the fence to the river are the y
components, then the perimeter is x + 2y =300. Therefore, x =300 - 2y. Now
since the area is just x times y, we can substitute 300 - 2y for x, and say
that the area (xy) is the same as (300 - 2y)*y, or area =300y - 2y^2. Now,
we could plot this out on a graph to see where the plot reaches a maximum
and determine the best values for the 2 sides of the fence and spend a
couple hours doing it. But since the first derivative of the function is the
rate of change, and the rate of change will be zero when the maximum area is
reached along the graph (since the area isn't getting any bigger), we can
use the first derivative of 300y - 2y ^2 which is 300 - 4y and set that to
0 so that 300 - 4y =0, or 4y =300, and y = 75 feet. And x is just the
remaining fence, or 150 feet. So the best fence dimensions would be 75 by
150 feet for a maximum area of 11,250 square feet of grass production.
news:2b430f0d-60ca-48eb-bfd7-00edf3b6cbdb@googlegroups.com...
The only calculus I remember is using the first derivative to find the rate
of change. An example is figuring out the optimum dimensions for a pot field
located on a river bank where you havea limited length of fence, say 300
feet. So, if the parallel part of the fence along the river is the x
component, and the perpenducular parts of the fence to the river are the y
components, then the perimeter is x + 2y =300. Therefore, x =300 - 2y. Now
since the area is just x times y, we can substitute 300 - 2y for x, and say
that the area (xy) is the same as (300 - 2y)*y, or area =300y - 2y^2. Now,
we could plot this out on a graph to see where the plot reaches a maximum
and determine the best values for the 2 sides of the fence and spend a
couple hours doing it. But since the first derivative of the function is the
rate of change, and the rate of change will be zero when the maximum area is
reached along the graph (since the area isn't getting any bigger), we can
use the first derivative of 300y - 2y ^2 which is 300 - 4y and set that to
0 so that 300 - 4y =0, or 4y =300, and y = 75 feet. And x is just the
remaining fence, or 150 feet. So the best fence dimensions would be 75 by
150 feet for a maximum area of 11,250 square feet of grass production.
