Find the local maximum and minimum values and saddle point(s) of the function. I

ID: 2873123 • Letter: F

Question

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

f(x, y) = 1 2x + 4y x2 4y2

2.–/1.25 pointsSCalc7 14.7.506.XP.My Notes

f(x, y) = x3y + 36x2 8y

local maximum value(s) local minimum value(s) saddle point(s) (x, y, f) =

Explanation / Answer

1. f(x,y)= 1-2x+4y-x2-4y2

fx = -2-2x; fy = 4-8y; fxx =-2 ; fyy =-8 and fxy =0

Equating fx=0 and fy =0 gives x= -1 and y= 1/2 which is the critical point.

Using second partial derivative test for determining local maxima/ minimum, consider the quantity,

d= fxx *fyy -[fxy]2 =(-2) (-8) -0 = 16, d>0 and also fxx =-2 <0

Hence there is a local maxima at (-1,1/2). This gives local maximum value f(-1,1/2)=3. Local minimum value DNE. There is no saddle point(DNE).

2. f(x,y)= x3y+36x2-8y

fx=3x2y+72x; fy= x3-8; fxx=6xy+72; fyy=0; fxy=3x2

Equating fx=0 and fy=0, gives critical point x=2,y=-12

Using second partial derivative test for determining local maxima,minima or saddle point consider the quantity,

d= fxx (2,-12) *fyy (2,-12) - [fxy (2, -12)]2 ; {fxx (2-12) = -72, fyy (2,-12)=0, fxy (2,-12) =12}

=(-72*0) - (12)2 = -144 which is <0

Hence saddle point is (2,-12, 144). There is no local maxima/minima(DNE).

(3D graphing facility is not available)

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Find the local maximum and minimum values and saddle point(s) of the function. I

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Find the local maximum and minimum values and saddle point(s) of the function. I

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