Question

find extreme values of f(x,y) = x^2 + 2y^2 on the disk x2 + y2 <=1 .. how???

Answer 1

The points INSIDE as well as the points ON the circle x^2 + y^2 = 1 represents all possible points. Which points will maximize f(x,y)? Which points will minimize f(x,y)?

Answer 2

f(x,y) = x^2 + 2y^2
Both terms are squared and so no matter what the value of (x,y) is, f(x,y) will always be greater than or equal to zero (that is, it cannot be negative).
f(x,y) >= 0

So the lowest value occurs when the point is the origin, (0,0) when f(x,y) = 0.

How about the points that maximizes f(x,y)?

Answer 3

Thanks!

Answer 4

\[
f(x,y) = x^2 + 2y^2 \\
\text{Step 1. Find the critical points:} \\
f_x = 2x = 0 \implies x = 0 \\
f_y = 4y = 0 \implies y = 0 \\
x = 0, y = 0 \text{ satisfies the constraint } x^2 + y^2 \le 1\\
(0,0) \text{ is the only critical point.} \\
\text{ } \\
\text{Step 2. Use Lagrange Multiplier. Treat constraint as an EQUALITY.} \\
f(x,y) = x^2 + 2y^2 \\
g(x,y) = x^2 + y^2 = 1 ~~ \text{ ------ (1) }\\
f_x = \lambda g_x \implies 2x = \lambda 2x, ~~2x(\lambda - 1) = 0
\implies x = 0 \text{ or } \lambda = 1\\
f_y = \lambda g_y \implies 4y = \lambda 2y, ~~2y(\lambda - 2) = 0
\implies y = 0 \text{ or } \lambda = 2\\
\]

Answer 5

\[
g(x,y) = x^2 + y^2 = 1 ~~\text{----- (1)} \\
\text{From previous reply: } x = 0 \text{ or } \lambda = 1
\text{ or } y = 0 \text{ or } \lambda = 2 \\
\text{ } \\

x = 0. ~~ \text{From equation (1): } ~~y = \pm 1. ~~\text{Thus, }(0, -1), (0,1) \\
\text{ } \\

\lambda = 1. ~~\text{Substitute in }4y = \lambda2y: \\
4y = 2y \implies y = 0. ~~\text{From equation (1): } ~~x = \pm 1.
~~\text{Thus, }(-1, 0), (1, 0) \\
\text{ } \\

\text{The other two possibilities, }y = 0 \text{ or }\lambda = 2, \text{ will give the same four points. }\\
\text{Substitute each of the four points and the critical point in f(x,y) and find max/min.}

\]