7. Lagrange multipliers.

In the figure below we have illustrated an extreme value problem with constraints. The point A is the largest value of the function z=f(x,y) while the point B is the largest value of the function under the constraintg(x,y)=0.

Theorem: (Lagrange multipliers) Let f and g be differntiable functions, where f’x(x0,y0) and f’y(x0,y0) not both are zero. If the function f(x,y) has an extreme value in the point (x0,y0) under the constraint g(x,y)=0, then there is a constant  such that 



Remark: A similar theorem holds for a general function f(x1,x2,…,xn) with m constraints.

Example: The case n=3, m=2. Write the constraints as g1(x,y,z)=0 and g2(x,y,z)=0 and consider


An (if existent) extreme value can be found by solving the system


Example 6: How to construct a box with the same shape as in the figure below with a fix volume V0 such that the amount of material used will be minimal?


Solution: The problem consists of minimizing the amount of material (proportional to the area) 


under the constraint 


By Lagrange’s multiplier method we therefore construct 

and then solve the system


Multiply the first equation with a, the second with b and the third with c. By using that abc=V0, we get

If we subtract the second equation from the first we get

since a,b,c>0. If we then put a=b in the last two equations we get 


Now subtract the second of these two equations from the first


We should thus choose 


Inserted in the expression for the volumen we then have


This imples that we should choose