The minimum value of #f(x,y)=x^2+13y^2-6xy-4y-2# is?

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Answer 1 · 2018-07-23T07:29:01

#f(x,y)=x^2+13y^2-6xy-4y-2#

#=>f(x,y)=x^2-2*x*(3y)+(3y)^2+(2y)^2-2*(2y)*1+1^2-3#
#=>f(x,y)=(x-3y)^2+(2y-1)^2-3#

Minimum value of each squared expression must be zero.

So #[f(x,y)]_"min"=-3#

Answer 2 · 2018-07-23T08:04:32

There is a relative minimum at #(3/2,1/2)# and #f(3/2,1/2)=-3#

I think that we must calculate the partial derivatives.

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Here,

#f(x,y)=x^2+13y^2-6xy-4y-2#

The first partial derivatives are

#(delf)/(delx)=2x-6y#

#(delf)/(dely)=26y-6x-4#

The critical points are

#{(2x-6y=0),(26y-6x-4=0):}#

#<=>#, #{(3y=x),(26y-6*3y-4=0):}#

#<=>#, #{(3y=x),(8y=4):}#

#<=>#, #{(x=3/2),(y=1/2):}#

The second partial derivatives are

#(del^2f)/(delx^2)=2#

#(del^2f)/(dely^2)=26#

#(del^2f)/(delxdely)=-6#

#(del^2f)/(delydelx)=-6#

The determinant of the Hessian matrix is

#D(x,y)=|((del^2f)/(delx^2),(del^2f)/(delxdely)),((del^2f)/(dely^2),(del^2f)/(delydelx))|#

#=|(2,-6),(-6,26)|#

#=52-36#

#=16>0#

As #D(x,y)>0#

and

#(del^2f)/(delx^2)=2>0#

There is a relative minimum at #(3/2,1/2)#

And

#f(3/2,1/2)=1.5^2+13*0.5^2-6*1.5*0.5-4*0.5-2=-3#