Mathematical Methods for Physics and Engineering - Matematica.NET

5.9 STATIONARY VALUES UNDER CONSTRAINTS◮The temperature of a point (x, y) on a unit circle is given by T (x, y) =1+xy. Findthetemperature of the two hottest points on the circle.We need to maximise T (x, y) subject to the constraint x 2 + y 2 = 1. Applying (5.27) and(5.28), we obtainy +2λx =0, (5.29)x +2λy =0. (5.30)These results, together with the original constraint x 2 + y 2 = 1, provide three simultaneousequations that may be solved for λ, x and y.From (5.29) and (5.30) we find λ = ±1/2, which in turn implies that y = ∓x. Rememberingthat x 2 + y 2 = 1, we find thaty = x ⇒ x = ± √ 1 ,2y = ± √ 12y = −x ⇒ x = ∓ √ 1 ,2y = ± √ 1 .2We have not yet determined which of these stationary points are maxima and which areminima. In this simple case, we need only substitute the four pairs of x- andy- values intoT (x, y) =1+xy to find that the maximum temperature on the unit circle is T max =3/2 atthe points y = 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 thatthe number of constraints is smaller than the number of variables. For example,if we wish to find the stationary points of f(x, y, z) subject to the constraintsg(x, y, z) =c 1 and h(x, y, z) =c 2 ,wherec 1 and c 2 are constants, then we proceedas above, obtaining∂(f + λg + µh) =∂f∂x ∂x + λ ∂g∂x + µ ∂h∂x = 0,∂(f + λg + µh) =∂f∂y ∂y + λ∂g ∂y + µ ∂h∂y= 0, (5.31)∂(f + λg + µh) =∂f∂z ∂z + λ∂g ∂z + µ∂h ∂z = 0.We may now solve these three equations, together with the two constraints, togive λ, µ, x, y and z.169