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y2-10-Sy + 1 (1 point) Suppose f(x,y) -x2 + (A) How many critical points does f

ID: 2888123 • Letter: Y

Question

y2-10-Sy + 1 (1 point) Suppose f(x,y) -x2 + (A) How many critical points does f have in R2? (B) If there is a local minimum, what is the value of the discriminant D at that point? If there is none, type N (C) If there is a local maximum, what is the value of the discriminant D at that point? If there is none, type N (D) If there is a saddle point, what is the value of the discriminant D at that point? If there is none, type N. at is the maximum value on (F) What is the minimum value of f on R2? If there is none, type N

Explanation / Answer

Given f(x,y) = x^2 + y^2 - 10 x - 8 y + 1

fx = /x ( x^2 + y^2 -10x - 8 y + 1 )                and           fy = /y ( x^2 + y^2 - 10x - 8y + 1

   = 2x - 10                                                                     = 2y - 8

fxx = /x ( 2x - 10 )                                     and           fyy = /y ( 2y - 8 )

    = 2                                                                                = 2

fxy = /y ( 2x - 10 )                                     and          fyx = /x ( 2y - 8 )

     = 0                                                                              = 0

for critical Points

fx = 0                                                          and    fy = 0

2 x - 10 = 0                                                                     2y - 8 = 0

2x = 10                                                                           2 y = 8

x = 5                                                                                  y = 4

so ( 5 , 4 ) is only critical Point

Second derivative Test :

D(a,b) = fxx(a,b) * fyy( a, b) - [ fxy(a,b) * fyx(a,b) ]

If D >0 and fxx(a, b) >0, then f(a, b) is a local minimum

If D >0 and fxx(a, b) < 0, then f(a, b) is a local maximum

If D < 0 , then (a, b) is a saddle point

D( 5 , 4 ) = fxx( 5 , 4 ) * fyy( 5 , 4 ) - [ fxy( 5 , 4 ) * fyx( 5 , 4 ) ]

             = 2 * 2 - [ 0 * 0 ]

             = 4 > 0

fxx( 5 , 4 ) = 2 > 0

so

f(5,4) = 5^2 + 4^2 - 10 (5) - 8 (4) + 1

        = 25 + 16 - 50 - 32 + 1

f(5,4) = - 40( minimum value ) is a local minimum

( A )

Critical Point is ( 5,4 )

( B )

Local minimum , D = 4

( C )

N

( D )

N

( E )

N

( F )

f (5,4) = - 40 ( minimum value )