Q. 34 - 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. $g(x,y)=x^3+6x^2+6y^2-4$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{local} \mathrm{minimum} \mathrm{at} \left(0,0\right) \mathrm{with} g(0,0)=-4 \mathrm{and} \\ \mathrm{saddle} \mathrm{point} \mathrm{at} (-4,0) \mathrm{with} g(-4,0)=28 \end{gathered}$$

Step 1. Given information

A function, $g(x,y)=x^3+6x^2+6y^2-4$

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

$$\begin{gathered} \mathrm{The} \mathrm{first}-\mathrm{order} \mathrm{partial} \mathrm{derivatives} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{are}: \\ {g}_x(x,y)=\frac{\partial g}{\partial \mathrm{x}}=3{\mathrm{x}}^2+12x \mathrm{and} g_y(x,y)=\frac{\partial g}{\partial \mathrm{y}}=12y \\ \mathrm{Now}, \mathrm{solve} \mathrm{the} \mathrm{system} \mathrm{of} \mathrm{equations}: 3{\mathrm{x}}^2+12x=0 \mathrm{and} 12y=0, \mathrm{we} \mathrm{get}, \\ x(3x+12)=0 \mathrm{and} y=0 \\ \Rightarrow x=0, x=-4 and y=0 \\ \mathrm{We} \mathrm{find} \mathrm{two} \mathrm{stationary} \mathrm{point}s \mathrm{of} g, \mathrm{namely}: (0,0), (-4,0) \\ \mathrm{The} \mathrm{second}-\mathrm{order} \mathrm{partial} \mathrm{derivatives} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{are}: \\ {g}_{xx}(x,y)=\frac{{\partial }^{2}g}{\partial {\mathrm{x}}^2}=6x+12, g_{yy}(x,y)=\frac{{\partial }^{2}g}{\partial {\mathrm{y}}^2}=12 \mathrm{and} g_{xy}(x,y)=\frac{{\partial }^{2}g}{\partial \mathrm{x}\partial \mathrm{y}}=0 \\ {g}_{xx}(0,0)=12, g_{yy}(0,0)=12 and g_{xy}(0,0)=0 \\ {g}_{xx}(-4,0)=-24, g_{yy}(-4,0)=12 and g_{xy}(-4,0)=0 \\ \mathrm{The} \mathrm{determinant} \mathrm{of} \mathrm{the} \mathrm{Hessian} \mathrm{is}: \\ det\left(H_g\left(x,y\right)\right)=\left(\frac{{\partial }^{2}g}{\partial {\mathrm{x}}^2}\right)\left(\frac{{\partial }^{2}g}{\partial {\mathrm{y}}^2}\right)-{\left(\frac{{\partial }^{2}g}{\partial \mathrm{x}\partial \mathrm{y}}\right)}^2 \\ det\left(H_g\left(0,0\right)\right)=12\times 12-0=144 \\ det\left(H_g\left(-4,0\right)\right)=-24\times 12-0=-288 \end{gathered}$$

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

$$\begin{gathered} \mathrm{If} g \mathrm{has} \mathrm{a} \mathrm{stationary} \mathrm{point} \mathrm{at} ({\mathrm{x}}_0,{\mathrm{y}}_0), \mathrm{then} \\ \left(\mathrm{a}\right) g \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{maximum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_g\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {g}_{xx}(x_0,y_0)<0 or g_{yy}(x_0,y_0)<0. \\ \left(\mathrm{b}\right) g \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{minimum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_g\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {g}_{xx}(x_0,y_0)>0 \mathrm{or} g_{yy}(x_0,y_0)>0. \\ \left(\mathrm{c}\right) g \mathrm{has} \mathrm{a} \mathrm{saddle} \mathrm{point} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_g\right(x_0,y_0\left)\right)<0. \\ \left(\mathrm{d}\right) \mathrm{If} det\left(H_g\right(x_0,y_0\left)\right)=0, \mathrm{no} \mathrm{conclusion} \mathrm{may} \mathrm{be} \mathrm{drawn} \mathrm{about} \mathrm{the} \mathrm{behavior} \mathrm{of} \\ g \mathrm{at} (x_0,y_0). \\ \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{function}, det\left(H_g\left(0,0\right)\right)=144>0 \mathrm{with} g_{xx}\left(0,0\right)=12>0 \\ \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{minimum} \mathrm{at} \left(0,0\right) \mathrm{with} \mathrm{minimum} \mathrm{value}, \\ g\left(0,0\right)=0^3+6{\left(0\right)}^2+6{\left(0\right)}^2-4=-4 \\ \mathrm{Also}, \det \left({\mathrm{H}}_{\mathrm{g}}\left(-4,0\right)\right)=-288<0 \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{saddle} \mathrm{point} \\ \mathrm{at} \left(-4,0\right) \mathrm{with} \mathrm{g}\left(-4,0\right)={\left(-4\right)}^3+6{\left(-4\right)}^2+6{\left(0\right)}^2-4=28 \end{gathered}$$

Step 4. Testing and finding absolute maximum and absolute minimum

$$\begin{gathered} \mathrm{When} y=0,\underset{x\to \infty }{\lim }f(x,0)=\infty \mathrm{and} \underset{x\to -\infty }{\lim }f(x,0)=-\infty \\ \mathrm{Therefore}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{local} \mathrm{minimum} \mathrm{at} \left(0,0\right) \mathrm{and} \mathrm{saddle} \mathrm{point} \mathrm{at} (-4,0) \end{gathered}$$