maxima, minima, quadratic approx.

[pk]

let f(x,y) = A(x^2)+Bxy+C(y^2)+Dx+Ey
where A,C >0 and B^2-4AC <0. Show that the minimum value of f(x,y) is
[BED-A(E^2)-C(D^2)]/[4AC-(B^2)]

[nooonuii]

ข้อความเดิมเขียนโดยคุณ pk

let

$f(x,y) = Ax^2+Bxy+Cy^2+Dx+Ey$

where $A,C >0$ and $B^2-4AC <0$. Show that the minimum value of $f(x,y)$ is

$\dfrac{BED-AE^2-CD^2}{4AC-B^2}$

First, set up the equation

$f_x=2Ax+By+D=0$

$f_y=2cy+Bx+E=0$

Then solve for $x,y$ by using Cramer's rule.

$x_0=\dfrac{BE-2CD}{4AC-B^2}$

$y_0=\dfrac{BD-2AE}{4AC-B^2}$.

Check that $(x_0,y_0)$ gives a minimum by using the second partial derivative test at $(x_0,y_0)$.

$f_{xx}=2A>0$

$f_{xy}=B$

$f_{yy}=2C$

$f_{xx}f_{yy}-(f_{xy})^2=4AC-B^2>0$.

Thus $f(x_0,y_0)$ is a minimum.