Very important, Lagrange multiplier

In summary: You're welcome.In summary, the conversation discusses a math problem involving finding the maximum and minimum values of a function on the boundary of a constraint. The person asks for help and is advised to review introductory calculus. The conversation then delves into using Lagrange multipliers to solve the problem, and ultimately finds the constrained max and min values. The expert provides a complete solution and the person expresses their gratitude for the help.
  • #1
odporko
6
0
Guys, i would be really greatfull if someone help me with this because i really don't know how to deal with this math problem: Find the maximum and minimum values of f = x^(1/4) + y^(1/3) on the boundary of g = 4*x+ 6*y = 720.
Please help me someone, i am desperate from this :(
 
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  • #2
odporko said:
Guys, i would be really greatfull if someone help me with this because i really don't know how to deal with this math problem: Find the maximum and minimum values of f = x^(1/4) + y^(1/3) on the boundary of g = 4*x+ 6*y = 720.
Please help me someone, i am desperate from this :(

You need to first show what you have done before we are allowed to offer help.

RGV
 
  • #3
L ( x , y ) = x^(1/4) + y^(1/3) + λ ( 4*x + 6*y − 720 )

I am not able to partialy derivate this equation, it always has weird numbers and after that i am not able to continue because i don't understand L. multiplier very well.
 
  • #4
odporko said:
L ( x , y ) = x^(1/4) + y^(1/3) + λ ( 4*x + 6*y − 720 )

I am not able to partialy derivate this equation, it always has weird numbers and after that i am not able to continue because i don't understand L. multiplier very well.

Are you saying you do not know how to differentiate x^(1/4) and x with respect x? If that is really true you need to go back and review introductory calculus.

RGV
 
  • #5
1. Homework Statement

f(x,y) = x^(1/4)*y^(1/3), g(x,y) = 4*x +6*y=720
sorry i had a mistake before

2. Homework Equations

∇f=λ∇g
y^(1/3)/(4*x^(3/4))= 4*λ
x^(1/4)/(3*y^(2/3))= 6*λ

these are partial derivations, and we want to express y just with lambda and here is something that really terrifies me: 4096*((104976*y^(8/3)*λ)^(9/4))*λ, am i thinking right? or am i going wrong?
 
  • #6
it goes for y=(1/ 812 479 653 374 328*λ^(13/4))^(1/5)... and now my mind already blew up
 
  • #7
4*(1/450868486864896*λ^(11/3))^(1/5)+ 6*(1/ 812479653374328*λ^(13/4))^(1/5)-720=0
well this is the equation after i get x and y and now i am pretty sure i did something wrong...
 
  • #8
odporko said:
1. Homework Statement

f(x,y) = x^(1/4)*y^(1/3), g(x,y) = 4*x +6*y=720
sorry i had a mistake before

2. Homework Equations

∇f=λ∇g
y^(1/3)/(4*x^(3/4))= 4*λ
x^(1/4)/(3*y^(2/3))= 6*λ
Good. So eliminate [itex]\lambda[/itex] by dividing the first equation by the second:
[tex]\frac{y^{1/3}}{4x^{3/4}}\frac{3y^{2/3}}{x^{1/4}}=\frac{3}{4}\frac{y}{x}= \frac{2}{3}[/tex]
so we must have y= (8/9)x. Put that into the equation of the constraint, 4x+6y= 720, to solve for x.

these are partial derivations, and we want to express y just with lambda and here is something that really terrifies me: 4096*((104976*y^(8/3)*λ)^(9/4))*λ, am i thinking right? or am i going wrong?
Well, what we want to do is to solve three equations in x, y, and [itex]\lambda[/itex] for x and y. We do not necessarily "want to express y just with lambda". There are many ways of solving for x and y. Since finding a value for [itex]\lambda[/itex] is not part of the solution, I suggest eliminating [itex]\lambda[/itex] first by dividing one equation by another.
 
  • #9
odporko said:
1. Homework Statement

f(x,y) = x^(1/4)*y^(1/3), g(x,y) = 4*x +6*y=720
sorry i had a mistake before

2. Homework Equations

∇f=λ∇g
y^(1/3)/(4*x^(3/4))= 4*λ
x^(1/4)/(3*y^(2/3))= 6*λ

these are partial derivations, and we want to express y just with lambda and here is something that really terrifies me: 4096*((104976*y^(8/3)*λ)^(9/4))*λ, am i thinking right? or am i going wrong?

You can use the first equation to solve for y in terms of (x,λ), then substitute that y into the second equation to get an equation containing only x and λ. Solve for x in terms of λ, then substitute that x into the expression for y. You will end up with x and y expressed only in terms of λ, and the expressions are not too horrible:
[tex] x = \frac{2^{4/5}\, 3^{2/5}}{2304\, \lambda^{12/5}}, \;
y = \frac{2^{4/5}\, 3^{2/5}}{2592\, \lambda^{12/5}}.[/tex]
Substituting these into the equation for g you get a simple linear equation in the variable
[itex] \lambda^{12/5}, [/itex] so you can solve it and pick the relevant real, positive root. This turns out to be
[tex] \lambda = \frac{5^{7/12}\, 3^{1/12}\, 7^{5/12}\, 2^{1/6}}{720} \doteq 0.009827792083.[/tex] Now you can find x and y:
[tex] x = 540/7, \; y = 480/7, \; f(x,y) = \frac{540^{1/4}\, 7^{5/12} \,480^{1/3}}{7}
\doteq 12.13030337.[/tex] This is the constrained max.

The constrained min cannot be found by Lagrange multipliers. Just recognize that f(x,y) is not defined for x < 0 and y < 0, so we are restricted to the region {x ≥ 0, y ≥ 0}, and that f = 0 when x = 0 or y = 0. Thus, there are two constrained min points, at (x,y) = (0,720/6) and (x,y) = (720/4,0), both giving f = 0.

RGV
 
  • #10
Thank you very much, you really really helped me probably pass the most important subject this semester :)
 
  • #11
An entire subject just on Lagrange Multipliers? Lol.
 

1. What is the Lagrange multiplier method and why is it important in scientific research?

The Lagrange multiplier method is a mathematical technique used to optimize a function subject to constraints. It is important in scientific research because it allows scientists to find the maximum or minimum value of a function while satisfying certain constraints, which is a common problem in many scientific fields.

2. How does the Lagrange multiplier method work?

The Lagrange multiplier method involves creating a new function, called the Lagrangian, by combining the original function with a multiple of the constraints. The solution to the original problem can then be found by finding the critical points of the Lagrangian function.

3. What are the assumptions of the Lagrange multiplier method?

The Lagrange multiplier method assumes that the original function and the constraints are continuous and differentiable. It also assumes that the constraints are independent of each other.

4. In what types of scientific problems is the Lagrange multiplier method commonly used?

The Lagrange multiplier method is commonly used in optimization problems, such as in economics, engineering, and physics. It is also used in mathematical modeling and data analysis.

5. What are the advantages and limitations of using the Lagrange multiplier method?

The advantages of the Lagrange multiplier method include its ability to handle multiple constraints and its efficiency in finding the optimal solution. However, it may not always find the global optimal solution and can be computationally expensive for complex problems.

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