Q. 35 - Calculus

Question

In Exercises 31–52, find the relative maxima, relative minima, and saddle points for the given functions. Determine whether the function has an absolute maximum or absolute minimum as well. $f(x,y)=x^3-12xy+y^3$

Step-by-Step Solution

Answer

There are no maximum, minimum points or saddle points.

Step 1. Given information

A function, $f(x,y)=x^3-12xy+y^3$

Step 2. Finding the first-order, second-order partial derivatives and determinant of hessian

$$\begin{gathered} \mathrm{The} \mathrm{first}-\mathrm{order} \mathrm{partial} \mathrm{derivatives} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{are}: \\ {f}_x(x,y)=\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=3{\mathrm{x}}^2-12y \mathrm{and} f_y(x,y)=\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=3{\mathrm{y}}^2-12x \\ \mathrm{Now}, \mathrm{solve} \mathrm{the} \mathrm{system} \mathrm{of} \mathrm{equations}: 3{\mathrm{x}}^2-12y=0 \mathrm{and} 3{\mathrm{y}}^2-12x=0, \mathrm{we} \mathrm{get}, \\ {\mathrm{x}}^2=4y \mathrm{and} {\mathrm{y}}^2=4x \\ \Rightarrow {\left(\frac{x^2}{4}\right)}^2=4x \\ \Rightarrow x\left(x^3-64\right)=0 \\ \Rightarrow x=0, y=0 \mathrm{and} x=4, y=4 \\ \mathrm{We} \mathrm{find} \mathrm{two} \mathrm{stationary} \mathrm{point}s \mathrm{of} f, \mathrm{namely}: (0,0), (4,4) \\ \mathrm{The} \mathrm{second}-\mathrm{order} \mathrm{partial} \mathrm{derivatives} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{are}: \\ {f}_{xx}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{x}}^2}=6x-12, f_{yy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}=6y-12 \mathrm{and} f_{xy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}=-12 \\ {f}_{xx}(0,0)=-12, f_{yy}(0,0)=-12 \mathrm{and} f_{xy}(0,0)=-12 \\ {f}_{xx}(4,4)=12, f_{yy}(4,4)=12 \mathrm{and} f_{xy}(4,4)=-12 \\ \mathrm{The} \mathrm{determinant} \mathrm{of} \mathrm{the} \mathrm{Hessian} \mathrm{is}: \\ det\left(H_f\left(x,y\right)\right)=\left(\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{x}}^2}\right)\left(\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}\right)-{\left(\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}\right)}^2 \\ det\left(H_f\left(0,0\right)\right)=\left(-12\right)\times \left(-12\right)-{\left(-12\right)}^2=0 \\ det\left(H_f\left(4,4\right)\right)=12\times 12-{\left(-12\right)}^2=0 \end{gathered}$$

Step 3. Testing and finding relative maximum, relative minimum and saddle points

$$\begin{gathered} \mathrm{If} f \mathrm{has} \mathrm{a} \mathrm{stationary} \mathrm{point} \mathrm{at} ({\mathrm{x}}_0,{\mathrm{y}}_0), \mathrm{then} \\ \left(\mathrm{a}\right) f \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{maximum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_f\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {f}_{xx}(x_0,y_0)<0 or f_{yy}(x_0,y_0)<0. \\ \left(\mathrm{b}\right) f \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{minimum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_f\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {f}_{xx}(x_0,y_0)>0 \mathrm{or} f_{yy}(x_0,y_0)>0. \\ \left(\mathrm{c}\right) f \mathrm{has} \mathrm{a} \mathrm{saddle} \mathrm{point} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_f\right(x_0,y_0\left)\right)<0. \\ \left(\mathrm{d}\right) \mathrm{If} det\left(H_f\right(x_0,y_0\left)\right)=0, \mathrm{no} \mathrm{conclusion} \mathrm{may} \mathrm{be} \mathrm{drawn} \mathrm{about} \mathrm{the} \mathrm{behavior} \mathrm{of} \\ f \mathrm{at} (x_0,y_0). \\ \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{function}, det\left(H_f\left(0,0\right)\right)=0, \mathrm{hence}, \mathrm{no} \mathrm{conclusion} \mathrm{can} \mathrm{be} \mathrm{made} \\ \mathrm{Also}, det\left(H_f\left(4,4\right)\right)=0, \mathrm{hence}, \mathrm{no} \mathrm{conclusion} \mathrm{can} \mathrm{be} \mathrm{made} \end{gathered}$$

Step 4. Testing and finding absolute maximum and absolute minimum

There are no maximum and minimum points.