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 where 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 