Answer to Question #108189 in Calculus for Garima Ahlawat

Question

(1) find the minimum value of the function f(x,y) = x^2 + 2y^2 on the circle x^2 + y^2 =1
(2) let the function f be defined by f(x,y) = 3x^2 y^4 / (x^4 + y^8) (x,y) not equal to (0,0) and 0 , (x,y) = (0,0). Show that f has directional derivatives in all directions at (0,0)

Answer 1

(1). Make the $L=(x,y,l)=x^2+2y^2+l(x^2+y^2-1)$ where $l$ is Lagrange maltiplier.

$L'_x=2x+2lx,L'_y=4y+2ly.$ Solving system $L'_x=L'_y=0$ and $x^2+y^2=1$ then

$l=-1,y=0,x=1,x=-1$

$L''_{x^2}=2+2l,L''_{xy}=0,L''_{y^2}=4+2l$ then $d^2L=(2+2l)dx^2+(4+2l)dy^2$ .If $l=-1$

then $d^2L>0$ hence $(1,0),(-1,0)$ is points of minimum $f(1,0)=f(-1,0)=1$

(2). $\lim_{x,y\to0} \frac{3x^2y^4}{x^4+y^8}=\lim_{x\to0}\frac{3k}{k^2+1}=\frac{3k}{k^2+1}$ where $k=\frac{x^2}{y^4}.$ The limit of the $f(x,y)$ depends

on the value of $k$ . Hence $f$ has directional derivatives

in all directions at $(0,0).$