Find the minimum value of $f(x, y) = x^2 + y^2 + 6x - 4y + 15$.
Completing the square: f(x, y) = (x+$3)^2 - 9 + (y-2)^2 - 4 + 15$ = (x+$3)^2 + (y-2)^2 + 2$. Minimum occurs at (-3, 2) with value 2.
For the function $f(x, y) = x^3 + y^3 - 3xy$, the point (1, 1) is a:
f_x = $3x^2 - 3y$, f_y = $3y^2 - 3x$. At (1,1): r = f_xx = 6, s = f_xy = -3, t = f_yy = 6. rt - $s^2$ = 36 - 9 = 27 > 0 and r = 6 > 0, so (1,1) is a local minimum.
For $f(x, y) = x^3 + y^3 - 3xy$, what is the nature of the critical point (0, 0)? Enter 1 for Local Min, 2 for Local Max, 3 for Saddle Point.
At (0,0): r = f_xx = 0, s = f_xy = -3, t = f_yy = 0. rt - $s^2$ = 0 - 9 = -9 < 0. Since rt - $s^2$ < 0, (0,0) is a saddle point (Type 3).
If u = $\ln\left(\frac{x^4 + y^4}{x + y}\right)$, then by Euler's Theorem, x $\frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ is equal to:
Let $f = e^u$ = $(x^4 + y^4)/(x + y)$. f is homogeneous of degree 4 - 1 = 3. By Euler's theorem, x f_x + y f_y = 3f. Since f_x = $e^u$ u_x, x $e^u$ u_x + y $e^u$ u_y = 3 $e^u$ => x u_x + y u_y = 3.
If $u = x^2$ y + $y^2$ z + $z^2$ x, evaluate $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}$ at (1, 1, 1).
u_x = 2xy + $z^2$. At (1,1,1) u_x = 3. u_y = $x^2 + 2yz$. At (1,1,1) u_y = 3. u_z = $y^2 + 2zx$. At (1,1,1) u_z = 3. Sum = 3 + 3 + 3 = 9.
