Lagrange Multipliers #2: Two Variables, 1 Constraint

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Lagrange Multipliers #2: Two Variables, 1 Constraint

Рет қаралды 21,333

Күн бұрын

In this video we go over how to use Lagrange Multipliers to find the absolute maximum and absolute minimum of a function given a constraint curve. Specifically we find the maximum and minimum values of f(x,y) = 4x^3+y^2 given 2x^2+y^2=1. If you're neat and organized you'll be fine!
Some other things you might hear while you do these: constrained optimization, absolute max/min over a curve.
This is a very common topic to cover in a Calculus III/Multivariable calculus course. Good luck!

Пікірлер: 11

@itsadam4316 2 ай бұрын

You're honestly a life safer. So clear with every step and you broke down something very daunting into a bunch of simple steps.

@williamramsfeldt201 6 ай бұрын

Thank you! Very clear example and easy to follow. Helped me with my own problem.

@turksvids 6 ай бұрын

Glad it helped! Lagrange multipliers is definitely one of the tricky topics you run into.

@jamessullivan9363 Жыл бұрын

thank you for a good example with such clear work

@b.f.skinner4383 4 жыл бұрын

Great video, easy to follow along!

@newtonocharimenyenya2458 3 жыл бұрын

A very Great piece.

@junior5146 3 жыл бұрын

I have a hard time with this. Sometimes you solve the system of eqs with respect to lambda and sometimes with x and y. I am not sure when to choose which. Can the lambda ever equal zero when calculating lagrange multipliers?

@turksvids 3 жыл бұрын

Yeah, the hard part of Lagrange Multipliers is really solving the system. For my purposes I almost never need to know the value of lambda so I usually try to solve equal equation for lambda and then set those equations equal to each other. Unfortunately there's not really a hard fast rule of what to do.

@anandabiswas5460 7 ай бұрын

How to apply this on functions like 3x^2 + 4xy + y^2. like when xy are together multiplied.....solving the system of equations then becomes a real challenge ! any idea ?