Lagrange multipliers: 2 constraints

[@jmoriarty974]

Putting 2x-y=0 made your question be very easy. If it was not there, how would we go about it

[@medderapper9061]

Thanks for your help 🙂

[@petripaavonpoika3767]

Suppose I have three interdependent functions: A, B and C. Then, I will write a Lagrangian L1 = A-l1((B-b)-l2(C-c) Extremizing the Lagrangian will give me the multipliers l1 and l2. But then, could I write two more Lagrangians L2=C-l3(A-a)-l4(B-b) L3=B-l5(C-c)-l6(A-a) Extremizing the latter two Lagrangians should give me the multipliers l3, l4, l5 and l6, right?

[@FB-tr2kf]

I get everything else, but cannot see how you got Lambda 2 to equal 4 lambda 1 @6:50.

[@henrikswedish378]

from 3 you get z =-0.5 lambda2 from C2 you get z = -2lambda1, since z=z, it follows, -0.5 lambda2= -2lambda1, divide both sides with -0.5 gives you lambda2 = 4 lambda1

[@anshimasingh7993]

How can we say that this will find maximum value of f? If we have to find minimum value of f then what will be the procedure.

[@ujjawalbarnwal2390]

I am from India people of India interested in math how people think about India in this matter like math please any foreigners reply Thank you