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 845
Example25.18 Locate and identifythe stationarypoints of
f(x,y)= x3 y2
−x+
3 2 +2y
Solution The first partialderivatives arefound:
∂f
∂x =x2 −1
∂f
∂y =y+2
The first partialderivatives areequated tozero:
x 2 −1=0
y+2=0
These equations have two solutions:x = 1,y = −2 andx = −1,y = −2.
In order to identify the nature of each stationary point, the second derivatives are
found:
∂ 2 f
∂x 2 = 2x
∂ 2 f
∂x∂y = 0
∂ 2 f
∂y 2 = 1
Each stationary point isexamined inturn.
At (1,−2)
( )
∂ 2 f ∂ 2 f ∂ 2
∂x 2 ∂y − f 2
=2x(1)−0 2 =2x=2sincex=1
2 ∂x∂y
∂ 2 f
∂x 2 =2x=2
Since both expressions are positive the point (1,−2) isaminimum point.
At (−1,−2)
Here
∂ 2 f ∂ 2 (
f ∂ 2
∂x 2 ∂y − f
2 ∂x∂y
) 2
=2x=−2sincex=−1
Since the expression isnegative then (−1,−2) isasaddle point.
Figure 25.2 illustratesaplot of the surface defined by f (x,y).
z
1
0.5
0
–0.5
–1
–1.5
–2
–2.5
–3
0
–1
–2
y
–3
–4 –2
–1
0
1
2
x
Figure25.2
Thefunction
f(x,y)= x3
3
−x+
y2
2 +2y
has a minimumat (1, −2)
and asaddle point at
(−1, −2).