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

844 Chapter 25 Functions of several variables
Example25.16 Locate and identify the stationary values of
f(x,y)=x 2 +xy+y
Solution The first partialderivatives arefound:
∂f
∂x =2x+y
∂f
∂y =x+1
The first partialderivatives areequated tozero:
2x+y=0
x+1=0
This yieldsx = −1,y = 2.Thus thereisone stationary point,positioned at (−1,2).
The second partialderivatives arefound:
∂ 2 f
∂x 2 = 2
∂ 2 f
∂x∂y = 1
∂ 2 f
∂y 2 = 0
Now
∂ 2 f ∂ 2 (
f ∂ 2 )
∂x 2 ∂y − f 2
=2(0) − (1) 2 = −1
2 ∂x∂y
Since the expression isnegative, we conclude that (−1,2) isasaddle point.
Example25.17 Locateand identifythe stationarypoints of
f(x,y)=xy−x 2 −y 2
Solution Thefirst partialderivatives arefound:
∂f
∂x =y−2x
∂f
∂y =x−2y
Solving ∂f
∂x
= 0,
∂f
∂y =0yieldsx=0,y=0.
The second partialderivatives arefound:
∂ 2 f
∂x 2 = −2
∂ 2 f
∂x∂y = 1
∂ 2 f
∂y 2 = −2
The second derivative testtoidentify the stationary point isused. Now
∂ 2 f ∂ 2 (
f ∂ 2 )
∂x 2 ∂y − f 2
= (−2)(−2) − (1) 2 =3
2 ∂x∂y
( )
Since ∂2 f f ∂ 2 f ∂ 2
∂x 2 <0and∂2 ∂x 2 ∂y − f 2
> 0 then (0, 0) isamaximum point.
2 ∂x∂y