Find the extremum of f(x,y) subject to the given constraint, and state whether it is a maximum or a minimum. f ( x,y )=2 x^2 +3y^2 - 4 xy; x+ y =27 There is a minimum value of 162 located at (x, y)=( 15,12 ).
Accepted Solution
A:
The Lagrangian is[tex]L(x,y,\lambda)=2x^2+3y^2-4xy+\lambda(x+y-27)[/tex]with critical points whenever[tex]L_x=4x-4y+\lambda=0\implies\lambda=4y-4x[/tex][tex]L_y=6y-4x+\lambda=0\implies\lambda=4x-6y[/tex][tex]L_\lambda=x+y-27=0[/tex]Solving for [tex]x[/tex] and [tex]y[/tex], we get[tex]L_x=L_y=0\implies4y-4x=4x-6y\implies5y=4x[/tex][tex]L_\lambda=0\implies x+\dfrac45x=\dfrac95x=27\implies x=15[/tex][tex]L_\lambda=0\implies15+y=27\implies y=12[/tex]so there is one critical point at (15, 12), at which point [tex]f(15,12)=162[/tex].The Hessian matrix for this function is[tex]H(x,y)=\begin{bmatrix}4&-4\\-4&6\end{bmatrix}[/tex]with determinant 8 > 0, and [tex]f_{xx}=4>0[/tex], which tells us this point is a minimum.