Use Lagrange Multipliers to Find the Maximum and Minimum Values of f(x,y) = x^3y^5

Use Lagrange Multipliers to Find the Maximum and Minimum Values of f(x,y) = x^3y^5 constrained to the line x+y=8/5.
To use Lagrange multipliers we always set up the equation grad(f) = L grad(g) then solve the nonlinear system.
The hardest part is generally solving the non linear system. The best thing you can do is try many different problems to get practice.
Thanks for watching! I hope Lagrange multipliers don't seem so intimidating!
-dr. dub

[@mathCS1233]

thank you so much!!!

[@JonathanWaltersDrDub]

No problem!

[@raynz1337]

what if the function value returns the same number for every critical point? what would be the maximum or minimum values if I do not get a second function value to compare. supposing the function value = 3 for every point.

[@FactsPolice]

sem question

[@user-dx1hv5nu8u]

you're still responding, that's nice
In the last step if I have less than 0, it's still a minimum?
And what if I got square equation and becouse of that got 2 points, I need to mark them as min/max separately?

[@JonathanWaltersDrDub]

In Lagrange multiplier problems the largest function value is the max and the smallest function value is the min (regardless of whether it’s positive or negative.

[@user-dx1hv5nu8u]

@@JonathanWaltersDrDub thank you

[@JonathanWaltersDrDub]

@@user-dx1hv5nu8u You're welcome!

[@geewanga4360]

Thanks much I have now understood

[@JonathanWaltersDrDub]

I’m so glad!

[@bibekmalik5224]

Hallo I am from India

[@germanml4432]

thanks!

[@JonathanWaltersDrDub]

Always happy to help!

[@gmail1565]

There is no video in this example. U are great

[@JonathanWaltersDrDub]

Thanks for your support!

[@gmail1565]

How u choose x=0, y=0 in 3x^2y^5=5x^3y^4

[@gmail1565]

I have doubt on this

[@quintile9705]

So it isn't necessary to do the D test because you know they are crit points and you can just plug them in?

[@JonathanWaltersDrDub]

It isn’t necessary to do D because any continuous function on a closed and bounded domain will achieve its absolute extrema somewhere.
This is analogous to optimizing f(x) on a closed interval [a,b].
We don’t classify the critical points inside we just compare all the function values.

[@AngmarCosplay]

In the beginning, why is 8/5 not in the g(x,y) equation?

[@JonathanWaltersDrDub]

Because g(x,y) is just the left hand side of the constraint equation. g(x,y) = c for some constant c is the constraint equation.

[@AngmarCosplay]

@@JonathanWaltersDrDub gotcha, thanks!

[@anthony_9263]

Thank you, I was in agony.

[@JonathanWaltersDrDub]

Glad I could help!

[@Elman219]

i need music

[@JonathanWaltersDrDub]

😂