Question

maximization problem
Build a rectangle pen with three parallel partitions
using 400 feet fencing. what dimensions will maximize the the total area.

Answer 1

Checking if I'm right!
we have P=2y+4x=400
y=200-2x
A=xy ====> A=x(200-2x)
A'=200-4x ===>200-4x=0 ===> x=50
y=200-100=100

Answer 2

I'm not sure about three parallel partitions thingy!
is the diagram correct

Answer 3

the diagram is fine

Answer 4

f = xy g = 4x + 2y - 400

using lagrange multiplier method: when the gradient of f and g are simply a multiple of each other, then we have critical points to play with

fx = y Lgx = 4L
fy = x Lgy = 2L

x = 2L
y = 4L

by substitution:
g = 0 = 4(2L) + 2(4L) - 400

400 = 16L, L = 25

therefore x = 50, and y = 100 :)

Answer 5

oh seems i was right!
i have no idea about Lagrange multiplier method
our professor didn't mention that in the course

Answer 6

didnt mention it in my course either ... had to go figure it out all on me own ;)

Answer 7

oh i see! good thing
any page, book to check that method i looks nice lol

Answer 8

none that come to mind off hand. pauls calculus site might be a good primer in it.

Answer 9

http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx

Answer 10

ok thanks, i will check that!
===================
just a question came to my mind, when i solved couple of optimization problem
i find solutions
when i check my answer, i find that they use some boundaries
like this example:
an open rectangle box with square base is to be constructed from 48 ft^2 of material. What dimensions will give the largest possible volume
when they solved
they said 0<x<= sqrt(48)
i know that x cannot be zero
how about sqrt part?

Answer 11

I found solution without those boundaries though

Answer 12

if we have a function and multiple constaints such as say:

f, constrained by g and h

then the gradient of f is equal to a linear combination of the gradient of the constraints.

\[\nabla f=\lambda \nabla g+ \kappa \nabla h\]

Answer 13

an open rectangle box with square base

is to be constructed from 48 ft^2 of material.

What dimensions will give the largest possible volume

when they solved they said 0<x<= sqrt(48)
------------------------------------

now the question is: do we use all 48 ft^2 or do we assume its a square shape of material and that there is cut offs?