Max and Min of a function of four variables

[Artie]

Homework Statement

Find the maximum and minimum values of the function f(x,y,z,t)=x+y+z+t subject to the constraint x^2+y^2+z^2+t^2=400.

Homework Equations

I think the Lagrange multiplers can be used ∇f=λ∇g

The Attempt at a Solution

So I found ∇f=<1,1,1,1> and ∇g=<2x,2y,2z,2t>
and when i set each component equal to each other I get x=y=z=t. I don't know where to go from here, or if this was even the right path to take in the first place

 

[Simon Bridge]

Would you prefer answers here or https://www.physicsforums.com/showthread.php?t=643816?
Have you tried considering the Lagrange function from these?
http://en.wikipedia.org/wiki/Lagrange_multiplier

 

[Artie]

Sorry, I forgot that there was a designated spot for homework questions. But yes I've looked all over the internet for how to do this problem. So what is the next step?
I know how to use Lagrange multipliers for just two variables, f(x,y), but I don't understand how to use it for four.

 

[Simon Bridge]

The next step is to construct the Lagrange equation.

The method is supposed to be useful for any number of independent variables.
http://www.karlscalculus.org/pdf/lagrange.pdf

 

[Artie]

okay so, for each component:
1=λ2x
1=λ2y
1=λ2z
1=λ2t

 

[Simon Bridge]

That's 4 equations and five unknowns.
You are missing one.

 

[Artie]

are you talking about lambda?
if so, i don't know an equation for that which doesn't include atleast one other variable

 

[Ray Vickson]

are you talking about lambda?
if so, i don't know an equation for that which doesn't include atleast one other variable

You can solve for the 4 variables x,y,z,t in terms of λ. So, if you know λ you are done. You still need to satisfy the constraint equation. Try substituting your expressions for x,y,z,t in the constraint to see what you get.

That is more-or-less the standard solution method for Lagrange multiplier problems, whether you have 2 variables or 2000 variables.

RGV

 

[Simon Bridge]

I know how to use Lagrange multipliers for just two variables

... in that method, you'd end up using three relations wouldn't you? Did you go look at either of the links I gave you? The second explicitly deals with the case of more than two variables.

 

[Artie]

Okay I think I got it!
Since all of the components are equal to 1, you can set them equal to each other. And you end up with x=y=z=t. So you plug that into the constraint function g(x,y,z,t)=400, substituting x for all the other values. And you get (x^2)+(x^2)+(x^2)+(x^2)=400. And from there you get x=+/-10. Which means that y,z, and t also equal +/-10. So you plug these values back into the function f(x,y,z,t) and you get a max value of 40 occurring at the point (10,10,10,10) and a min value of -40 occurring at the point (-10,-10,-10,-10)
Right?

 

[Artie]

Okay I think I got it!
Since all of the components are equal to 1, you can set them equal to each other. And you end up with x=y=z=t. So you plug that into the constraint function g(x,y,z,t)=400, substituting x for all the other values. And you get (x^2)+(x^2)+(x^2)+(x^2)=400. And from there you get x=+/-10. Which means that y,z, and t also equal +/-10. So you plug these values back into the function f(x,y,z,t) and you get a max value of 40 occurring at the point (10,10,10,10) and a min value of -40 occurring at the point (-10,-10,-10,-10)
Right?

 

[Artie]

Okay I think I got it!
Since all of the components are equal to 1, you can set them equal to each other. And you end up with x=y=z=t. So you plug that into the constraint function g(x,y,z,t)=400, substituting x for all the other values. And you get (x^2)+(x^2)+(x^2)+(x^2)=400. And from there you get x=+/-10. Which means that y,z, and t also equal +/-10. So you plug these values back into the function f(x,y,z,t) and you get a max value of 40 occurring at the point (10,10,10,10) and a min value of -40 occurring at the point (-10,-10,-10,-10)
Right?