# Multivariable Calculus: Finding the minimum and maximum value

1. Jun 11, 2012

### theBEAST

1. The problem statement, all variables and given/known data
Given f(x,y) = xy(x+2y-6), let D be the region in the plane between the hyperbola xy = 4 (let this be g) and the line x+2y-6 = 0 (let this be h). Find the maximum and minimum values of f(x,y) on D.

3. The attempt at a solution
I first found the critical points for f(x,y) and it turns out that none of them are in this region. Then I used lagrange to find the critical points on the boundary:
https://dl.dropbox.com/u/64325990/Photobook/Photo%202012-06-11%207%2026%2054%20PM.jpg [Broken]

Solving this by hand and also confirming with wolfram I get the critical points (2,2) and (4,1). HOWEVER according to the answer key, there is one more critical point at (2√2,2/√2). They did this by solving the boundary case for xy=4 in between the regions 2 and 4 as shown here:
https://dl.dropbox.com/u/64325990/Capture.PNG [Broken]

I am wondering why the Lagrange method I used did not give me (2√2,2/√2) as well?

Last edited by a moderator: May 6, 2017
2. Jun 11, 2012

### Vargo

You are using the Lagrange method that applies when you have two simultaneous constraints. So you are looking for the "critical points" that happen at the intersection of your hyperbola and your line.

You need to use Lagrange with one constraint at a time. Notice that the critical point you missed lies on the hyperbola but not the line. Or you could use substitution and reduce to a calc 1 problem, but that is up to you.

3. Jun 12, 2012

### HallsofIvy

Staff Emeritus
In other words, you can use Lagrange multipliers to find max and min values on xy= 4, then use a different Lagrange multiplier calculation to find max and min values on x+ 2y- 6= 0.

Or, you could simply put the "conditions" into the function itself. On xy= 4, y= 4/x so f(x,y)= xy(x+ 2y- 6)= 4(x+ 4/x- 6). On x+ 2y- 6= 0, it is even easier: f(x,y)= xy(0)= 0 for all x.