How do I determine the parameter a of the following equation such that f(x, y) has an extremum at a point (x, y) with x = -3, and show that it is a minimum?

#f(x,y)=ax^2+xy+y^2+x+y+1#

#f(x,y)=ax^2+xy+y^2+x+y+1#

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Mar 8, 2018

Answer:

See below.

Explanation:

Calling

#p = (x,y)#
#p_0 = (-3,1)#
#a=1/3#

we have

#f(x,y) = (p-p_0) cdot A cdot (p-p_0) #

with

#A = ((1/3,1/2),(1/2,1))#

#f(x,y)# has a minimum at #p = p_0#

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