The minimum value of f(x,y)=x^2+13y^2-6xy-4y-2f(x,y)=x2+13y2−6xy−4y−2 is?

2 Answers
P dilip_k
Jul 23, 2018

f(x,y)=x^2+13y^2-6xy-4y-2f(x,y)=x2+13y2−6xy−4y−2

=>f(x,y)=x^2-2*x*(3y)+(3y)^2+(2y)^2-2*(2y)*1+1^2-3⇒f(x,y)=x2−2⋅x⋅(3y)+(3y)2+(2y)2−2⋅(2y)⋅1+12−3
=>f(x,y)=(x-3y)^2+(2y-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[f(x,y)]min=−3

Narad T.
Jul 23, 2018

There is a relative minimum at (3/2,1/2)(32,12) and f(3/2,1/2)=-3f(32,12)=−3

Explanation:

I think that we must calculate the partial derivatives.

![slideplayer.com](useruploads.socratic.orguseruploads.socratic.org)

Here,

f(x,y)=x^2+13y^2-6xy-4y-2f(x,y)=x2+13y2−6xy−4y−2

The first partial derivatives are

(delf)/(delx)=2x-6y∂f∂x=2x−6y

(delf)/(dely)=26y-6x-4∂f∂y=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

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