Q. 33 - 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. $f(x,y)=x^3+y^3-12x-3y+15$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{a} \mathrm{local} \mathrm{minimum}: f(2,1)=-3, \mathrm{a} \mathrm{local} \mathrm{maximum}: \\ f(-2,-1)=33 \mathrm{and} \mathrm{saddle} \mathrm{points}: f(2,-1)=1, f(-2,1)=29 \end{gathered}$$

Step 1. Given information

A function, $f(x,y)=x^3+y^3-12x-3y+15$

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}}^2-12 \mathrm{and} f_y(x,y)=\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=3{\mathrm{y}}^2-3 \\ \mathrm{Now}, \mathrm{solve} \mathrm{the} \mathrm{system} \mathrm{of} \mathrm{equations}: 3{\mathrm{x}}^2-12=0 \mathrm{and} 3{\mathrm{y}}^2-3=0, \mathrm{we} \mathrm{get}, \\ {\mathrm{x}}^2=4 \mathrm{and} {\mathrm{y}}^2=1 \\ \Rightarrow x=\pm 2 \mathrm{and} y=\pm 1 \\ \mathrm{We} \mathrm{find} \mathrm{the} \mathrm{four} \mathrm{stationary} \mathrm{point}s \mathrm{of} f, \mathrm{namely}: (2,1), (2,-1), (-2,1),(-2,-1) \\ \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, f_{yy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}=6y \mathrm{and} f_{xy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}=0 \\ {f}_{xx}(2,1)=12, f_{yy}(2,1)=6 \mathrm{and} f_{xy}(2,1)=0 \\ {f}_{xx}(2,-1)=12, f_{yy}(2,-1)=-6 \mathrm{and} f_{xy}(2,-1)=0 \\ {f}_{xx}(-2,1)=-12, f_{yy}(-2,1)=6 \mathrm{and} f_{xy}(-2,1)=0 \\ {f}_{xx}(-2,-1)=-12, f_{yy}(-2,-1)=-6 \mathrm{and} f_{xy}(-2,-1)=0 \\ \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(2,1\right)\right)=12\times 6-0=72 \\ det\left(H_f\left(2,-1\right)\right)=12\times \left(-6\right)-0=-72 \\ det\left(H_f\left(-2,1\right)\right)=\left(-12\right)\times 6-0=-72 \\ det\left(H_f\left(-2,-1\right)\right)=\left(-12\right)\times \left(-6\right)-0=72 \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(2,1\right)\right)=72>0 \mathrm{with} f_{xx}\left(2,1\right)=12>0 \mathrm{and} f_{yy}\left(2,1\right)=6>0. \\ \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{minimum} \mathrm{at} \left(2,1\right) \mathrm{with} \mathrm{minimum} \mathrm{value}, \\ f\left(2,1\right)={\left(2\right)}^3+{\left(1\right)}^3-12\left(2\right)-3\left(1\right)+15=-3 \\ \mathrm{Also}, det\left(H_f\left(2,-1\right)\right)=-72<0. \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{a} \mathrm{saddle} \mathrm{point} \mathrm{at} \left(2,-1\right) \\ \mathrm{with} f(2,-1)={\left(2\right)}^3+{\left(-1\right)}^3-12\left(2\right)-3(-1)+15=1 \\ \mathrm{Also}, det\left(H_f\left(-2,1\right)\right)=-72<0. \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{a} \mathrm{saddle} \mathrm{point} \mathrm{at} \left(-2,1\right) \\ \mathrm{with} f(-2,1)= {\left(-2\right)}^3+{\left(1\right)}^3-12(-2)-3\left(1\right)+15=29 \\ Also, det\left(H_f\left(-2,-1\right)\right)=72>0 \mathrm{with} f_{xx}\left(-2,-1\right)=-12<0 \mathrm{and} f_{yy}\left(-2,-1\right)=-6<0. \\ \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{maximum} \mathrm{at} \left(-2,-1\right) \mathrm{with} \mathrm{maximum} \mathrm{value}, \\ f\left(-2,-1\right)={\left(-2\right)}^3+{\left(-1\right)}^3-12(-2)-3(-1)+15=33 \end{gathered}$$

Step 4. Testing and finding absolute maximum and absolute minimum

$$\begin{gathered} \mathrm{When} x=0,\underset{y\to \infty }{\lim }f(0,y)=\infty \mathrm{and} \underset{y\to -\infty }{\lim }f(0,y)=-\infty \\ \mathrm{Therefore}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{local} \mathrm{minimum} \mathrm{at} \left(2,1\right) \mathrm{and} \mathrm{local} \mathrm{maximum} \mathrm{at} (-2,-1) \end{gathered}$$