LaGrange max/min (calc 3)

[jaredmt]

Homework Statement

i needed to find the max and min possible volume for a box with edges that = 200cm and surface area that = 1500cm^2 using Lagrange multipliers.

Homework Equations

edges: 4x + 4y + 4z = 200cm
Area: 2xy + 2xz + 2yz = 1500 cm^2
Volume = xyz

The Attempt at a Solution

i brought it down to this:
Vol = f
Area = G
Edges = g

^f = ^G# + ^g$
(# and & just means 'some number')
( ^ is the gradient of the function)
when i simplify everything out i get these statements:

yz = #(y + z) + $
xz = #(x + z) + $
xy = #(x + y) + $
x + y +z = 50
xy + yz + yz = 750when i try to replace values it seems to get messy. do i just throw in random values for # and $ or something? I am 99% sure that they cannot = 0. that is pretty much all i can say from these statements right now lol

 

[Dick]

Let's call L=# and M=$. Subtract your first two equation from each other. This gives (x-y)*z=L*(x-y). Or (z-L)*(x-y)=0. This tells you either z=L or x=y. Similarly y=L or x=z, or x=L or y=z. If you think about it, in any case, two of the lengths must be equal. Put this information into your last two equations. Now you have two equations in two unknowns. Not hard to solve at all.

 

[jaredmt]

ok cool, thanks a lot, found the answer