Q. 38 - 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. $g(x,y)=8x^3-3x{y}^2+2y^3-4x-2$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{local} \mathrm{minimum} \mathrm{at} (0.43, 0.43) \mathrm{with} g(0.43, 0.43)=-3.16 \\ \mathrm{and} \mathrm{local} \mathrm{maximum} \mathrm{at} (-0.43, -0.43) \mathrm{with} g(-0.43, -0.43)=-4 \end{gathered}$$

Step 1. Given information

A function, $g(x,y)=8x^3-3x{y}^2+2y^3-4x-2$

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

$$\begin{gathered} \mathrm{The} \mathrm{first}-\mathrm{order} \mathrm{partial} \mathrm{derivatives} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{are}: \\ {g}_x(x,y)=\frac{\partial g}{\partial \mathrm{x}}=24{\mathrm{x}}^2-3y^2-4 \mathrm{and} g_y(x,y)=\frac{\partial g}{\partial \mathrm{y}}=-6xy+6y^2 \\ \mathrm{Now}, \mathrm{solve} \mathrm{the} \mathrm{system} \mathrm{of} \mathrm{equations}: 24{\mathrm{x}}^2-3y^2-4=0 \mathrm{and} -6xy+6y^2=0, \mathrm{we} \mathrm{get}, \\ x=y \mathrm{and} 21x^2=4 \\ \Rightarrow x=y=\pm \frac{2}{\sqrt{21}}=\pm 0.43 \\ \mathrm{We} \mathrm{find} \mathrm{one} \mathrm{stationary} \mathrm{point} \mathrm{of} g, \mathrm{namely}: (0.43, 0.43) \\ \mathrm{The} \mathrm{second}-\mathrm{order} \mathrm{partial} \mathrm{derivatives} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{are}: \\ {g}_{xx}(x,y)=\frac{{\partial }^{2}g}{\partial {\mathrm{x}}^2}=48x, g_{yy}(x,y)=\frac{{\partial }^{2}g}{\partial {\mathrm{y}}^2}=12y-6x \mathrm{and} g_{xy}(x,y)=\frac{{\partial }^{2}g}{\partial \mathrm{x}\partial \mathrm{y}}=-6y \\ {g}_{xx}(0.43, 0.43)=20.64, g_{yy}(0.43, 0.43)=2.58 \mathrm{and} g_{xy}(0.43, 0.43)=-2.58 \\ {g}_{xx}(-0.43, -0.43)=-20.64, g_{yy}(-0.43, -0.43)=-2.58 \mathrm{and} g_{xy}(-0.43, -0.43)=2.58 \\ \mathrm{The} \mathrm{determinant} \mathrm{of} \mathrm{the} \mathrm{Hessian} \mathrm{is}: \\ det\left(H_g\left(x,y\right)\right)=\left(\frac{{\partial }^{2}g}{\partial {\mathrm{x}}^2}\right)\left(\frac{{\partial }^{2}g}{\partial {\mathrm{y}}^2}\right)-{\left(\frac{{\partial }^{2}g}{\partial \mathrm{x}\partial \mathrm{y}}\right)}^2 \\ det\left(H_g(0.43, 0.43)\right)=20.64\times 2.58-{\left(-2.58\right)}^2=46.6 \\ det\left(H_g(-0.43, -0.43)\right)=\left(-20.64\right)\times \left(-2.58\right)-{\left(2.58\right)}^2=46.6 \end{gathered}$$

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

$$\begin{gathered} \mathrm{If} g \mathrm{has} \mathrm{a} \mathrm{stationary} \mathrm{point} \mathrm{at} ({\mathrm{x}}_0,{\mathrm{y}}_0), \mathrm{then} \\ \left(\mathrm{a}\right) g \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{maximum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_g\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {g}_{xx}(x_0,y_0)<0 or g_{yy}(x_0,y_0)<0. \\ \left(\mathrm{b}\right) g \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{minimum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_g\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {g}_{xx}(x_0,y_0)>0 \mathrm{or} g_{yy}(x_0,y_0)>0. \\ \left(\mathrm{c}\right) g \mathrm{has} \mathrm{a} \mathrm{saddle} \mathrm{point} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_g\right(x_0,y_0\left)\right)<0. \\ \left(\mathrm{d}\right) \mathrm{If} det\left(H_g\right(x_0,y_0\left)\right)=0, \mathrm{no} \mathrm{conclusion} \mathrm{may} \mathrm{be} \mathrm{drawn} \mathrm{about} \mathrm{the} \mathrm{behavior} \mathrm{of} \\ g \mathrm{at} (x_0,y_0). \\ \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{function}, det\left(H_g(0.43, 0.43)\right)>0 \mathrm{with} g_{xx}(0.43, 0.43)>0 \\ \mathrm{and} g_{yy}(0.43, 0.43)>0. \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{minimum} \mathrm{at} (0.43, 0.43) \\ \mathrm{with}, g(0.43, 0.43)=8{\left(0.43\right)}^3-3(0.43){\left(0.43\right)}^2+2{\left(0.43\right)}^3-4(0.43)-2=-3.16 \\ \mathrm{Also}, det\left(H_g(-0.43, -0.43)\right)>0 \mathrm{with} g_{xx}(-0.43, -0.43)<0 \\ \mathrm{and} g_{yy}(-0.43, -0.43)<0. \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{maximum} \mathrm{at} (-0.43, -0.43) \\ \mathrm{with}, g(-0.43, -0.43)=8{\left(-0.43\right)}^3-3(-0.43){\left(-0.43\right)}^2+2{\left(-0.43\right)}^3-4(-0.43)-2=-4 \end{gathered}$$

Step 4. Testing and finding absolute maximum and absolute minimum

$$\begin{gathered} \mathrm{When} y=0,\underset{x\to \infty }{\lim }f(x,0)=\infty \mathrm{and} \underset{x\to -\infty }{\lim }f(x,0)=-\infty \\ \mathrm{Therefore}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{local} \mathrm{minimum} \mathrm{at} (0.43, 0.43) \\ \mathrm{and} \mathrm{local} \mathrm{maximum} \mathrm{at} (-0.43, -0.43) \end{gathered}$$