Question

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 ).

Answer 1

The Lagrangian is

$L(x,y,\lambda)=2x^2+3y^2-4xy+\lambda(x+y-27)$

with critical points whenever

$L_x=4x-4y+\lambda=0\implies\lambda=4y-4x$

$L_y=6y-4x+\lambda=0\implies\lambda=4x-6y$

$L_\lambda=x+y-27=0$

Solving for $x$ and $y$, we get

$L_x=L_y=0\implies4y-4x=4x-6y\implies5y=4x$

$L_\lambda=0\implies x+\dfrac45x=\dfrac95x=27\implies x=15$

$L_\lambda=0\implies15+y=27\implies y=12$

so there is one critical point at (15, 12), at which point $f(15,12)=162$.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}4&-4\\-4&6\end{bmatrix}$

with determinant 8 > 0, and $f_{xx}=4>0$, which tells us this point is a minimum.