MIN and MAX for funtions of two and three variables with constrainments

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SUMMARY

The discussion focuses on solving optimization problems involving functions of two and three variables with constraints, specifically using the method of Lagrange multipliers. The user successfully derived the gradient vectors and set up the equations but expressed confusion regarding the relationship between the variables x, y, and lambda in determining minimum and maximum values. The discussion emphasizes the importance of checking algebraic manipulations and systematically analyzing cases to eliminate variables and find critical points for optimization.

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  • Understanding of Lagrange multipliers for constrained optimization
  • Familiarity with gradient vectors and their significance in optimization
  • Basic algebra skills for manipulating equations
  • Knowledge of critical point analysis in multivariable calculus
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  • Learn how to derive and interpret gradient vectors in optimization problems
  • Practice solving optimization problems with multiple constraints
  • Explore case analysis techniques in multivariable calculus
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Students and educators in calculus, particularly those focusing on optimization techniques, as well as anyone seeking to understand the application of Lagrange multipliers in real-world problems.

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Homework Statement


see problem 4 and 5 attachment


Homework Equations





The Attempt at a Solution


see problem 4 attachement
I found the gradient vectors of each and set fgrad=lamdbda*ggrad and used the constrainment equation to solve for all three variables. What is confusing me is I'm not sure what to do with the x,y,lambda values. How do they relate to the minimum and maximum values? I cannot attachemt my problem 5 attempt because it is too large of a file. but they are essential the same type of problem with different equations so if someone could please help me on number four i'll probably figure out number five. Also, can you check my that my algebra is correct in solving x,y,lambda for problem 4. Thank you!
 

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i need help asap
 
first, if 2xy^4=L*2x then either case#1 x=0 or case#2 x does not =0. if not, then you can solve L=y^4.
put this into the next equation to eliminate L. check both cases.

second, if 4x^2y^3=L*4y then AGAIN, case#3 y=0 or case#4 y does not =0.

eliminate L and put all these cases into the last equation which will probably give multiple solutions for each case. once you have a point x=? y=? put this into F(x,y)=? which one is smallest/largest?
 

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