082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

25.7 Maximum and minimum points of a function of two variables 843
Example25.15 Locate the stationary points of
f(x,y)= x3
3 +3x2 +xy+ y2
2 +6y
Solution The first partialderivatives arefound:
∂f
∂x =x2 +6x+y
∂f
∂y =x+y+6
These derivatives are equated tozero:
x 2 +6x+y=0 (25.1)
x+y+6=0 (25.2)
Equations (25.1) and (25.2) are solved simultaneously. From (25.2)y = −x − 6, and
substituting this into Equation (25.1) yields x 2 + 5x − 6 = 0. Solving this quadratic
equationgivesx = 1, −6.Whenx = 1,y = −7,andwhenx = −6,y = 0.Thefunction
has stationary values at (1,−7) and (−6,0).
Equating the first partial derivatives to zero locates the stationary points, but does not
identifythemasmaximumpoints,minimumpointsorsaddlepoints.To distinguishbetween
these various points a test involving second partial derivatives must be made.
Note that this is similar to locating and identifying turning points of a function of one
variable.
To identifythe stationarypoints weconsider the expression
( )
∂ 2 f ∂ 2 f ∂ 2
∂x 2 ∂y − f 2
2 ∂x∂y
If the expression isnegative atastationary point,then thatpoint isasaddle point.
Iftheexpressionispositiveandinaddition ∂2 f
∂x ispositiveatastationarypoint,then
thatpoint isaminimum point.
2
Iftheexpressionispositiveandinaddition ∂2 f
∂x isnegativeatastationarypoint,then
thatpoint isamaximum point.
2
If the expression is zero then further tests are required. These are beyond the scope
of the book.
Insummary:
∂ 2 f ∂ 2 (
f ∂ 2 )
∂x 2 ∂y − f 2
< 0 Saddle point
2 ∂x∂y
( )
∂ 2 f ∂ 2 f ∂ 2
∂x 2 ∂y − f 2
>0 2 ∂x∂y
and
∂ 2 f ∂ 2 (
f ∂ 2 )
∂x 2 ∂y − f 2
>0 2 ∂x∂y
and
∂ 2 f
∂x 2 > 0
∂ 2 f
∂x 2 < 0
Minimum point
Maximum point