PARTIAL DIFFERENTIATION
Find the stationary points off(x, y, z)=x^3 +y^3 +z^3 subject to the following constraints:
(i)g(x, y, z)=x^2 +y^2 +z^2 =1;
(ii)g(x, y, z)=x^2 +y^2 +z^2 =1andh(x, y, z)=x+y+z=0.Case (i). Since there is only one constraint in this case, we need only introduce a single
Lagrange multiplier to obtain
∂
∂x(f+λg)=3x^2 +2λx=0,∂
∂y(f+λg)=3y^2 +2λy=0, (5.32)∂
∂z(f+λg)=3z^2 +2λz=0.These equations are highly symmetrical and clearly have the solutionx=y=z=− 2 λ/3.
Using the constraintx^2 +y^2 +z^2 = 1 we findλ=±
√
3 /2 and so stationary points occur
at
x=y=z=±1
√
3
. (5.33)
In solving the three equations (5.32) in this way, however, we have implicitly assumed
thatx,yandzare non-zero. However, it is clear from (5.32) that any of these values can
equal zero, with the exception of the casex=y=z= 0 since this is prohibited by the
constraintx^2 +y^2 +z^2 = 1. We must consider the other cases separately.
Ifx= 0, for example, we require
3 y^2 +2λy=0,
3 z^2 +2λz=0,
y^2 +z^2 =1.
Clearly, we requireλ= 0, otherwise these equations are inconsistent. If neitherynor
zis zero we findy=− 2 λ/3=zandfromthethirdequationwerequirey=z=
± 1 /
√
- Ify=0,however,thenz=±1 and, similarly, ifz=0theny=±1. Thus the
stationary points havingx=0are(0, 0 ,±1), (0,± 1 ,0) and (0,± 1 /
√
2 ,± 1 /
√
2). A similar
procedure can be followed for the casesy=0andz= 0 respectively and, in addition
to those already obtained, we find the stationary points (± 1 , 0 ,0), (± 1 /
√
2 , 0 ,± 1 /
√
2) and(± 1 /
√
2 ,± 1 /
√
2 ,0).
Case (ii). We now have two constraints and must therefore introduce two Lagrange
multipliers to obtain (cf. (5.31))
∂
∂x(f+λg+μh)=3x^2 +2λx+μ=0, (5.34)∂
∂y(f+λg+μh)=3y^2 +2λy+μ=0, (5.35)∂
∂z(f+λg+μh)=3z^2 +2λz+μ=0. (5.36)These equations are again highly symmetrical and the simplest way to proceed is to
subtract (5.35) from (5.34) to obtain
3(x^2 −y^2 )+2λ(x−y)=0
⇒ 3(x+y)(x−y)+2λ(x−y)=0. (5.37)This equation is clearly satisfied ifx=y; then, from the second constraint,x+y+z=0,