Solving Lagrange Multipliers: Max/Min f(x,y)

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SUMMARY

The discussion focuses on solving the problem of finding the maximum and minimum values of the function f(x,y) = x^3y under the constraint 3x^4 + y^4 = 1 using Lagrange multipliers. The solution involves calculating the gradients and setting up the equations ∇f = λ∇g, leading to λ = ±1/4 and the critical points x = y = ±1/√2. The maximum value of the function is confirmed to be 1/4, while the minimum value is -1/4, with two distinct solutions for both extrema.

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  • Understanding of Lagrange multipliers
  • Knowledge of gradient vectors
  • Familiarity with constraint optimization
  • Basic calculus and algebra skills
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Homework Statement


Using Lagrange multipliers, find the maximum and minimum values of [tex]f(x,y)=x^3y[/tex] with the constraint [tex]3x^4+y^4=1[/tex].

Homework Equations


The Attempt at a Solution


Here is my complete solution. I just wanted to make sure there are no errors and I did it correctly. Thanks for any feedback.

[tex]\nabla f = \lambda \nabla g[/tex]
[tex]3x^2y=12\lambda x^3[/tex] and [tex]x^3=4\lambda y^3[/tex]
Solving these I got [tex]\lambda = \frac{1}{4}[/tex] and [tex]\lambda = -\frac{1}{4}[/tex]

Putting these values into the equation on the right gives x=y and x=-y. Substituting these into the left equation gives [tex]x=y=\frac{1}{\sqrt{2}}[/tex] and [tex]x=\frac{1}{\sqrt{2}}[/tex], [tex]y = -\frac{1}{\sqrt{2}}[/tex].

Putting these values into the equation for [tex]f[/tex] gives a maximum of [tex]\frac{1}{4}[/tex] and minimum of [tex]-\frac{1}{4}[/tex].
 
Last edited:
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Your maximum and minimum values are correct, but you have, for the maxima two soluitons,
[tex]x=y=\pm\frac{1}{\sqrt{2}}[/tex]
rather then just one maximum.

Similarly for the minimum value.
 

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