Extremums
Local extremums definition
\displaylines{ \text{Note: further reading - Sylvester’s criterion} \ }
\displaylines{ f(x, y) = 3(x^{2} + y^{2}) + x^{3} + 4y \ \text{Find critical points of } f \text{ and categorize them} \ \ \text{Solution:} \ \nabla f = \begin{pmatrix} f_{x} \ f_{y} \ \end{pmatrix} = \begin{pmatrix} 6x + 3x^{2} \ 6y + 4 \ \end{pmatrix} \ \nabla f = 0 \iff \left{\begin{array}{} \left[\begin{array}{} x = 0 \ x = -2 \ \end{array}\right. \ y = -\frac{2}{3} \ \end{array}\right. \iff \left[\begin{array}{} \left( 0, -\frac{2}{3} \right) \ \left( -2, -\frac{2}{3} \right) \ \end{array}\right. \ H_{f} = \begin{pmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \ \end{pmatrix} = \begin{pmatrix} 6x + 6 & 0 \ 0 & 6 \ \end{pmatrix} \ H_{f}\left( 0, -\frac{2}{3} \right) = \begin{pmatrix} 6 & 0 \ 0 & 6 \ \end{pmatrix} \implies \boxed{ \left( 0, -\frac{2}{3} \right) \text{ is a local minimum} } \ H_{f}\left( -2, -\frac{2}{3} \right) = \begin{pmatrix} -6 & 0 \ 0 & 6 \ \end{pmatrix} \implies \boxed{ \left( -2, -\frac{2}{3} \right) \text{ is a saddle} } \ }
\displaylines{
- & \text{Saddle is a point where in some paths it is a local maximum and in some a minimum} \ & \text{It is named after the form of saddle, or Pringles} \
- & \text{For functions of one variable, there is a connection between the number of maximums} \ & \text{and minimums, between any two must be an opposite} \ & \text{For functions of multiple variable there aren’t} \ }
\displaylines{ f(x, y, z) = x^{2} + y^{2} + z^{2} - xy + yz - xz - 4x + 6y + 2z \ \nabla f = \begin{pmatrix} 2x - y - z - 4 \ 2y - x + z + 6 \ 2z + y - x + 2 \ \end{pmatrix} \ \nabla f = 0 \iff \left{\begin{array}{} 2x - y - z - 4 = 0 \ -x + 2y + z + 6 = 0 \ -x + y + 2z + 2 = 0 \ \end{array}\right. \iff \left{\begin{array}{} x + y + 2 = 0 \ -x + 2y + z + 6 = 0 \ -x + y + 2z + 2 = 0 \ \end{array}\right. \ \iff \left{\begin{array}{} y = -2 -x \ -x + 2y + z + 6 = 0 \ -x + y + 2z + 2 = 0 \ \end{array}\right. \iff \left{\begin{array}{} y = -2 - x \ x = 1 \ z = x \ \end{array}\right. \iff \left{\begin{array}{} x = 1 \ y = -3 \ z = 1 \ \end{array}\right. \ H_{f} = \begin{pmatrix} f_{xx} & f_{xy} & f_{xz} \ f_{yx} & f_{yy} & f_{yz} \ f_{zx} & f_{zy} & f_{zz} \ \end{pmatrix} = \begin{pmatrix} 2 & -1 & -1 \ -1 & 2 & 1 \ -1 & 1 & 2 \ \end{pmatrix} \ \det(M_{1}) = 2 > 0 \ \det(M_{2}) = 3 > 0 \ \det(M_{3}) = 4 > 0 \ }
\displaylines{ f(x, y, z) = x^{3} + y^{3} + z^{3} - 3xz - 3yz - 3xy \ \ \nabla f = \begin{pmatrix} 3x^{2} - 3z - 3y \ 3y^{2} - 3z - 3x \ 3z^{2} - 3x - 3y \ \end{pmatrix} \ \nabla = 0 \iff \left{\begin{array}{} x^{2} - z - y = 0 \ y^{2} - z - x = 0 \ z^{2} - x - y = 0 \ \end{array}\right. \iff \left{\begin{array}{} x^{2} - z - y = 0 \ y^{2} - x^{2} + y - x = 0 \ z^{2} - x - y = 0 \ \end{array}\right. \ \iff \left{\begin{array}{} x^{2} - z - y = 0 \ (y-x)(y+x+1) = 0 \ z^{2} - x - y = 0 \ \end{array}\right. \ y = x \implies \left{\begin{array}{} x^{2} - z - x = 0 \ z^{2} - x - x = 0 \ \end{array}\right. \implies \left{\begin{array}{} 2x = z^{2} \ x^{2} - z - x = 0 \ \end{array}\right. \ \implies \left{\begin{array}{} x = \frac{z^{2}}{2} \ \frac{z^{4}}{4} - \frac{z^{2}}{2} - z = z\left( z^{3} - 2z - 4 \right) = 0 \ \end{array}\right. \implies \left{\begin{array}{} x = \frac{z^{2}}{2} \ z(z-2)(z^{2} + 2z + 2) = 0 \ \end{array}\right. \ \implies \boxed{ \left[\begin{array}{} x = 0, y = 0, z = 0 \ x = 2, y = 2, z = 2 \ \end{array}\right. } \ y = - x - 1 \implies \left{\begin{array}{} x^{2} - z + x + 1 = 0 \ z^{2} - x + x + 1 = 0 \ \end{array}\right. \implies \left{\begin{array}{} x^{2} - z + x + 1 = 0 \ z^{2} = -1 \ \end{array}\right. \implies \text{No solutions} \ \ H_{f} = \begin{pmatrix} f_{xx} & f_{xy} & f_{xz} \ f_{yx} & f_{yy} & f_{yz} \ f_{zx} & f_{zy} & f_{zz} \ \end{pmatrix} = \begin{pmatrix} 6x & -3 & -3 \ -3 & 6y & -3 \ -3 & -3 & 6z \ \end{pmatrix} \ \det(M_{1}) = 6x \ \det(M_{2}) = 36xy - 9 \ \det(M_{3}) = 216xyz + 27 + 27 - 54y - 54z - 54x \ (0, 0, 0) \implies \det(M_{2}) < 0 \implies \boxed{ (0, 0, 0) \text{ is a saddle} } \ \boxed{ (2, 2, 2) \text{ is a local minimum} } \ }