Q. 39 - 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. $f(x,y)=8xy+\frac{1}{x}+\frac{1}{y}$

Step-by-Step Solution

Answer

$T\mathrm{he} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{local} \mathrm{minimum} \mathrm{at} \left(\frac{1}{2},\frac{1}{2}\right) \mathrm{with} f\left(\frac{1}{2},\frac{1}{2}\right) =6.$

Step 1. Given information

A function, $f(x,y)=8xy+\frac{1}{x}+\frac{1}{y}$

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}}=8y-\frac{1}{x^2} \mathrm{and} f_y(x,y)=\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=8x-\frac{1}{y^2} \\ \mathrm{Now}, \mathrm{solve} \mathrm{the} \mathrm{system} \mathrm{of} \mathrm{equations}: 8y-\frac{1}{x^2}=0 \mathrm{and} 8x-\frac{1}{y^2}=0, \mathrm{we} \mathrm{get}, \\ 8x=\frac{1}{{\left({\displaystyle \frac{1}{8x^2}}\right)}^2} \mathrm{and} y=\frac{1}{8x^2} \\ \Rightarrow 64x^4-8x=0 \\ \Rightarrow 8x(8x^3-1)=0 \\ \Rightarrow x=0 \mathrm{and} x=\frac{1}{2} \mathrm{but}, x \mathrm{and} y \mathrm{are} \mathrm{not} \mathrm{equal} \mathrm{to} \mathrm{zero} \\ \Rightarrow \mathrm{x}=\frac{1}{2}, \mathrm{y}=\frac{1}{2} \\ \mathrm{We} \mathrm{find} \mathrm{only} \mathrm{one} \mathrm{stationary} \mathrm{point} \mathrm{of} f, \mathrm{namely}: \left(\frac{1}{2},\frac{1}{2}\right) \\ \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}=\frac{2}{x^3}, f_{yy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}=\frac{2}{y^3} \mathrm{and} f_{xy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}=8 \\ {f}_{xx}\left(\frac{1}{2},\frac{1}{2}\right)=16, f_{yy}\left(\frac{1}{2},\frac{1}{2}\right)=16 \mathrm{and} f_{xy}\left(\frac{1}{2},\frac{1}{2}\right)=8 \\ \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(\frac{1}{2},\frac{1}{2}\right)\right)=16\times 16-{\left(8\right)}^2=192 \end{gathered}$$

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

$$\begin{gathered} \mathrm{If} f \mathrm{has} \mathrm{a} \mathrm{stationary} \mathrm{point} \mathrm{at} ({\mathrm{x}}_0,{\mathrm{y}}_0), \mathrm{then} \\ \left(\mathrm{a}\right) f \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{maximum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_f\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {f}_{xx}(x_0,y_0)<0 or f_{yy}(x_0,y_0)<0. \\ \left(\mathrm{b}\right) f \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{minimum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_f\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {f}_{xx}(x_0,y_0)>0 \mathrm{or} f_{yy}(x_0,y_0)>0. \\ \left(\mathrm{c}\right) f \mathrm{has} \mathrm{a} \mathrm{saddle} \mathrm{point} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_f\right(x_0,y_0\left)\right)<0. \\ \left(\mathrm{d}\right) \mathrm{If} det\left(H_f\right(x_0,y_0\left)\right)=0, \mathrm{no} \mathrm{conclusion} \mathrm{may} \mathrm{be} \mathrm{drawn} \mathrm{about} \mathrm{the} \mathrm{behavior} \mathrm{of} \\ f \mathrm{at} (x_0,y_0). \\ \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{function}, det\left(H_f\left(\frac{1}{2},\frac{1}{2}\right)\right)=192>0 \mathrm{with} f_{xx}\left(\frac{1}{2},\frac{1}{2}\right)=16>0 \\ \mathrm{and} f_{yy}\left(\frac{1}{2},\frac{1}{2}\right)=16>0. \\ \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{minimum} \mathrm{at} \left(\frac{1}{2},\frac{1}{2}\right) \mathrm{with} \mathrm{minimum} \mathrm{value}, \\ f\left(\frac{1}{2},\frac{1}{2}\right)=8\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\frac{1}{\left({\displaystyle \frac{1}{2}}\right)}+\frac{1}{{\displaystyle \left(\frac{1}{2}\right)}}=6 \end{gathered}$$

Step 4. Testing and finding absolute maximum and absolute minimum

$$\begin{gathered} \mathrm{When} x>0\underset{y\to -\infty }{\lim }f(x,y)=-\infty \\ \mathrm{Therefore}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{local} \mathrm{minimum} \mathrm{at} \left(\frac{1}{2},\frac{1}{2}\right) \mathrm{with} f\left(\frac{1}{2},\frac{1}{2}\right) =6 \end{gathered}$$