\[f(x,y)=x^2 + 4y + y^2 -9 y + 3xy\]


f1<-function(x,y) x^2 + 4*y + y^2 -9*y + 3*x*y


1) Take the partial derivatives.


library(Deriv)
## Warning: package 'Deriv' was built under R version 4.0.5
Deriv(f1)
## function (x, y) 
## c(x = 2 * x + 3 * y, y = 2 * y + 3 * x - 5)


\[f_x\prime(x,y) \ = \ 2x + 3y\]


\[f_y\prime(x,y) \ = \ 2y + 3x -5\]


2) Solve prime x for 0.

\[x=-\frac{3y}{2} \ and \ y= \frac{-2x}{3}\]

3) Plug above into the other partial derivative to isolate y=2 and thus the below point is a root in both derivatives.


(-3,2) (15/11,10/11)

f1(-3,2)
## [1] -15


If you walk up the the x axis only, you see the points are a minimum but only on the x axis.


f1(-2,2)
f1(-3,2)
f1(-4,2)
## [1] -14
## [1] -15
## [1] -14


Note the rules for the second derivative test

If D(x,y) > 0 and $f_xx(x,y) > 0 then Relative Minimum

If D(x,y) > 0 and $f_xx(x,y) < 0 then Relative Maximum

If D(x,y) < 0 and $f_xx(x,y) > 0 then Saddle Point

If D(x,y) = 0 then inconclusive.


\[D(x,y) \ = \ f_x\prime(x,y) * f_y\prime(x,y) - f_{xy}\prime\prime(x,y)f_{yx}\prime\prime(x,y)\]


Deriv(Deriv(f1))
## function (x, y) 
## c(x.x = 2, x.y = 3, y.x = 3, y.y = 2)


D = -9 so the test is inconclusive.



4) Display the function at various angles


x = seq(-10, 10)
y = seq(-10, 10)
z <- outer(x, y, f1)


for (i in 1:3) {
 theta<-10*i
 persp(x,y, z, theta = theta, phi = 0, expand = 0.5, col = "lightblue", main="x^2 + 4*y + y^2 -9*y + 3*x*y",axes=T, ticktype="detailed")

}