Using the Lagrange multiplier to find extrema?

  • #1
Hey guys! I have been on the forum for about a week or so and have compiled a lot of information and techniques to help me understand calculus, so i really appreciate everyone's help!

I am a soon-to-be freshman in college and am taking a summer class, calculus II (took calc I in HS). This is our last week of class after our final exam so my professor is taking this time to give us a preview of what we will be learning in the fall semester in Calc III (since this is the same professor). Every Tuesday class our professor gives us a few problems from future sections and asks us to "see what we can come up with" and to work together to find solutions. The following Tuesday he asks us to discuss the problems as a class, seeing which ones of us know our stuff =P

Basically, i want to ask you guys what you think about these problems as i do them along before i have my discussion. I really want to make a lasting impression on my professor by "knowing my stuff" -to show him i can do it! All's i need is a little help! Would you guys mind giving me some help?

We are using the textbook Calculus 8th edition by Larson, Hostetler and Edwards and the problems come from the book.

The problem is on pg 974 in chapter 13.10 in the text, number 10. It reads:

Use Lagrange multipliers to find the indicated extrema, assuming that x and y are positive.
And it gives:
minimize: f(x,y)= sqrt(x^2+y^2) with the constraint: 2x+4y-15=0

I looked at similar problems in the same section but they're little help. I did the best i could so far and this is what i came up with.

subject to the constraint:
2x+4y=15
let g(x,y)=2x+4y=15

since tri f(x,y)=x/sqrt(x^2+y^2)
I got this step, then in the next step i'm supposed to use some symbol that looks like, well i remember it as wavelength in physics, kinda looks like a k??
and tri g(x,y)=2(wavelength symbol)i+4(wavelength symbol)j

but to attain a system of linear equations, how do i do that with x/sqrt(x^2+y^2)??

Any help would be greatly appreciated. Thanks guys ;)
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
Where did you get i and j from? There aren't any vectors here. You have [tex]F = \sqrt{x^2 + y^2} + \lambda \phi[/tex] where [tex]\phi[/tex] denotes the constraint function g(x,y) = const.

The technique for this is:
Find [tex]\frac{\partial F}{\partial y} = 0[/tex]
[tex]\frac{\partial F}{\partial x} = 0[/tex]

and with the constraint equation you now have three equations and three unknowns x,y,[tex]\lambda[/tex]. Solve the system and verify the values.
 
  • #3
378
2
since tri f(x,y) is not =x/sqrt(x^2+y^2)
but

tri f(x,y) =(x/sqrt(x^2+y^2), y/sqrt(x^2+y^2)) = or say tri f

now tri f = lamda * tri g

g(x,y) = 2x+4y-15, so find tri g


tri g(x,y)=2(wavelength symbol)i+4(wavelength symbol)j
yes, you found it here (I don't think there should be lamda)

lamda comes there (00--> tri f = lamda * tri g)
this is one equation

and second is 2x+4y-15 = 0

hint : solve the #1 equation first (tri f = lamda * tri g)
.... left side i's = right side i's , same with j's

P.S. I love that larson book :!!) [used it last year]
 
Last edited:
  • #4
Thanks for that help!
I dont have time this sec to show all the work, but i'm getting

f( 15/8, 15,4) = sqrt( (15/8)^2 + (15/4)^2 ) = 15 sqrt(5) / 8 as my min

Can anyone confirm if this is correct? Thanks!
 
  • #5
378
2
I got y = 2x from the lambda eqn
and I substituted it to 2x+4y-15 = 0
and got different answer
x = 3/2 y = 3
 
  • #6
152
0
I'll let you figure out the Lagrange multiplier problem. However, I will say that I concur with rootX's solutions. I did it the old fashioned way. I solved 2x+4y-15=0 for y. Plugged into f(x,y) to turn it into a function of just x. Did the single variable min/max thing and got x=3/2.
 

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