MAXIMA, MINIMA AND SADDLE POINTS FOR FUNCTIONS OF TWO VARIABLES 357G
(iii) solve the simultaneous equations∂z
∂x=0 and
∂z
∂y=0 forxandy, which gives the co-ordinatesof the stationary points,(iv) determine∂^2 z
∂x^2,∂^2 z
∂y^2and∂^2 z
∂x∂y
(v) for each of the co-ordinates of the station-
ary points, substitute values ofxandyinto
∂^2 z
∂x^2,∂^2 z
∂y^2and∂^2 z
∂x∂yand evaluate each,(vi) evaluate(
∂^2 z
∂x∂y) 2
for each stationary point,(vii) substitute the values of∂^2 z
∂x^2,∂^2 z
∂y^2and∂^2 z
∂x∂yintothe equation=(
∂^2 z
∂x∂y) 2
−(
∂^2 z
∂x^2)(
∂^2 z
∂y^2)and evaluate,(viii) (a) if
0 then the stationary point is a
saddle point
(b) if
< 0 and∂^2 z
∂x^2< 0 , then the stationary
point is amaximum point,and(c) if
< 0 and∂^2 z
∂x^2> 0 , then the stationary
point is aminimum point36.4 Worked problems on maxima,
minima and saddle points for
functions of two variablesProblem 1. Show that the function
z=(x−1)^2 +(y−2)^2 has one stationary point
only and determine its nature. Sketch the surface
represented byzand produce a contour map in
thex-yplane.Following the above procedure:(i)∂z
∂x=2(x−1) and∂z
∂y=2(y−2)(ii) 2(x−1)= 0 (1)
2(y−2)= 0 (2)(iii) From equations (1) and (2),x=1 andy=2,
thus the only stationary point exists at (1, 2).(iv) Since∂z
∂x=2(x−1)= 2 x−2,∂^2 z
∂x^2= 2and since∂z
∂y=2(y−2)= 2 y−4,∂^2 z
∂y^2= 2and∂^2 z
∂x∂y=∂
∂x(
∂z
∂y)
=∂
∂x(2y−4)= 0(v)∂^2 z
∂x^2=∂^2 z
∂y^2=2 and∂^2 z
∂x∂y= 0(vi)(
∂^2 z
∂x∂y) 2
= 0(vii) =(0)^2 −(2)(2)=− 4(viii) Since<0 and∂^2 z
∂x^2>0,the stationary point
(1, 2) is a minimum.The surfacez=(x−1)^2 +(y−2)^2 is shown in three
dimensions in Fig. 36.7. Looking down towards the
x-yplane from above, it is possible to produce acon-
tour map. A contour is a line on a map which gives
places having the same vertical height above a datum
line (usually the mean sea-level on a geographical
map). A contour map forz=(x−1)^2 +(y−2)^2 is
shown in Fig. 36.8. The values ofzare shown on the
map and these give an indication of the rise and fall
to a stationary point.1o1
2xyzFigure 36.7