5.9 STATIONARY VALUES UNDER CONSTRAINTS
The temperature of a point(x, y)on a unit circle is given byT(x, y)=1+xy.Findthe
temperature of the two hottest points on the circle.We need to maximiseT(x, y) subject to the constraintx^2 +y^2 = 1. Applying (5.27) and
(5.28), we obtain
y+2λx=0, (5.29)x+2λy=0. (5.30)These results, together with the original constraintx^2 +y^2 = 1, provide three simultaneous
equations that may be solved forλ,xandy.
From (5.29) and (5.30) we findλ=± 1 /2, which in turn implies thaty=∓x.Remem-
bering thatx^2 +y^2 = 1, we find that
y=x ⇒ x=±1
√
2
,y=±1
√
2
y=−x ⇒ x=∓1
√
2
,y=±1
√
2
.
We have not yet determined which of these stationary points are maxima and which are
minima. In this simple case, we need only substitute the four pairs ofx-andy- values into
T(x, y)=1+xyto find that the maximum temperature on the unit circle isTmax=3/2at
the pointsy=x=± 1 /
√
2.
The method of Lagrange multipliers can be used to find the stationary points offunctions of more than two variables, subject to several constraints, provided that
the number of constraints is smaller than the number of variables. For example,
if we wish to find the stationary points off(x, y, z) subject to the constraints
g(x, y, z)=c 1 andh(x, y, z)=c 2 ,wherec 1 andc 2 are constants, then we proceed
as above, obtaining
∂
∂x(f+λg+μh)=∂f
∂x+λ∂g
∂x+μ∂h
∂x=0,∂
∂y(f+λg+μh)=∂f
∂y+λ∂g
∂y+μ∂h
∂y=0, (5.31)∂
∂z(f+λg+μh)=∂f
∂z+λ∂g
∂z+μ∂h
∂z=0.We may now solve these three equations, together with the two constraints, to
giveλ,μ,x,yandz.