Find the minimum value of f(x, y, z) = x^2 + 2y^2 + 3z^2 subject to the constraint x + 2y + 3z = 10

maximum-value
constraint

#1

Find the minimum value of f(x, y, z) = x^2 + 2y^2 + 3z^2 subject to the constraint x + 2y + 3z = 10. Show that f has no maximum value with this constraint.

Answer:

We have to minimize f(x,y,z) with respect to the condition that x+2y+3z-10=0. We write the Laplace function as:- L = f(x,y,z) - 位 (x+2y+3z-10);
1.饾浛L/ 饾浛x = 0 implies 2x -位 (1) = 0 implies 位 = 2x.
2. 饾浛L/ 饾浛y = 0 implies 4y -位 (2) = 0 implies 位 = 2y.
3. 饾浛L/ 饾浛z = 0 implies 6z -位 (3) = 0 implies 位 = 2z.
4. 饾浛L/ 饾浛位 =0 implies x+2y+3z = 10.
位 = 2x = 2y = 2z implies x=y=z. Substituting this in equation 4. implies,
x+2x+3x = 10
6x =10
x = 5/3 =y =z
minimum value of f(x,y,z) = (5/3)^2 + 2(5/3)^2 + 3(5/3)^2 = 6*25/9 = 50/3.
Now the constraint can be attained by decreasing any variable in negative direction and increasing others in positive, as the magnitude of x,y or z increases(in either positive or negative direction) f(x,y,z) increases since it contains only square terms. Therefore, because of no constraint on individual range of x,yor z, maximum value of f(x,y,z) does not exist.