Q. 37 - 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. $f(x,y)=2x^2-2xy+4y^2+3x+9y-5$

Step-by-Step Solution

Answer

$\mathrm{The} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{absolute} \mathrm{minimum} \mathrm{at} \left(-\frac{3}{2},-\frac{3}{2}\right) \mathrm{with} f\left(-\frac{3}{2},-\frac{3}{2}\right) =-14$

Step 1. Given information

A function, $f(x,y)=2x^2-2xy+4y^2+3x+9y-5$

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

$$\begin{gathered} \mathrm{The} \mathrm{first}-\mathrm{order} \mathrm{partial} \mathrm{derivatives} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{are}: \\ {f}_x(x,y)=\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=4x-2y+3 \mathrm{and} f_y(x,y)=\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=-2x+8y+9 \\ \mathrm{Now}, \mathrm{solve} \mathrm{the} \mathrm{system} \mathrm{of} \mathrm{equations}: 4x-2y+3=0 \mathrm{and} -2x+8y+9=0, \mathrm{we} \mathrm{get}, \\ \Rightarrow x=-\frac{3}{2} \mathrm{and} y=-\frac{3}{2} \\ \mathrm{We} \mathrm{find} \mathrm{only} \mathrm{one} \mathrm{stationary} \mathrm{point}s \mathrm{of} f, \mathrm{namely}: \left(-\frac{3}{2},-\frac{3}{2}\right) \\ \mathrm{The} \mathrm{second}-\mathrm{order} \mathrm{partial} \mathrm{derivatives} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{are}: \\ {f}_{xx}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{x}}^2}=4, f_{yy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}=8 \mathrm{and} f_{xy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}=-2 \\ {f}_{xx}\left(-\frac{3}{2},-\frac{3}{2}\right)=4, f_{yy}\left(-\frac{3}{2},-\frac{3}{2}\right)=8 \mathrm{and} f_{xy}\left(-\frac{3}{2},-\frac{3}{2}\right)=-2 \\ \mathrm{The} \mathrm{determinant} \mathrm{of} \mathrm{the} \mathrm{Hessian} \mathrm{is}: \\ det\left(H_f\left(x,y\right)\right)=\left(\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{x}}^2}\right)\left(\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}\right)-{\left(\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}\right)}^2 \\ det\left(H_f\left(-\frac{3}{2},-\frac{3}{2}\right)\right)=4\times 8-{\left(-2\right)}^2=28 \end{gathered}$$

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

$$\begin{gathered} \mathrm{If} f \mathrm{has} \mathrm{a} \mathrm{stationary} \mathrm{point} \mathrm{at} ({\mathrm{x}}_0,{\mathrm{y}}_0), \mathrm{then} \\ \left(\mathrm{a}\right) f \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{maximum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_f\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {f}_{xx}(x_0,y_0)<0 or f_{yy}(x_0,y_0)<0. \\ \left(\mathrm{b}\right) f \mathrm{has} \mathrm{a} \mathrm{relative} \mathrm{minimum} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_f\right(x_0,y_0\left)\right)>0 \mathrm{with} \\ {f}_{xx}(x_0,y_0)>0 \mathrm{or} f_{yy}(x_0,y_0)>0. \\ \left(\mathrm{c}\right) f \mathrm{has} \mathrm{a} \mathrm{saddle} \mathrm{point} \mathrm{at} (x_0,y_0) \mathrm{if} det\left(H_f\right(x_0,y_0\left)\right)<0. \\ \left(\mathrm{d}\right) \mathrm{If} det\left(H_f\right(x_0,y_0\left)\right)=0, \mathrm{no} \mathrm{conclusion} \mathrm{may} \mathrm{be} \mathrm{drawn} \mathrm{about} \mathrm{the} \mathrm{behavior} \mathrm{of} \\ f \mathrm{at} (x_0,y_0). \\ \mathrm{In} \mathrm{the} \mathrm{given} \mathrm{function}, det\left(H_f\left(-\frac{3}{2},-\frac{3}{2}\right)\right)=28>0 \mathrm{with} f_{xx}\left(-\frac{3}{2},-\frac{3}{2}\right)=4>0 \\ \mathrm{and} f_{yy}\left(-\frac{3}{2},-\frac{3}{2}\right)=8>0. \\ \mathrm{Hence}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{minimum} \mathrm{at} \left(-\frac{3}{2},-\frac{3}{2}\right) \mathrm{with} \mathrm{minimum} \mathrm{value}, \\ f\left(-\frac{3}{2},-\frac{3}{2}\right)=2{\left(-\frac{3}{2}\right)}^2-2\left(-\frac{3}{2}\right)\left(-\frac{3}{2}\right)+4{\left(-\frac{3}{2}\right)}^2+3\left(-\frac{3}{2}\right)+9\left(-\frac{3}{2}\right)-5=-14 \end{gathered}$$

Step 4. Testing and finding absolute maximum and absolute minimum

$$\begin{gathered} \mathrm{When} y=0,\underset{x\to \infty }{\lim }f(x,0)=\infty \mathrm{and} \underset{x\to -\infty }{\lim }f(x,0)=\infty \\ \mathrm{Therefore}, \mathrm{the} \mathrm{given} \mathrm{function} \mathrm{has} \mathrm{absolute} \mathrm{minimum} \mathrm{at} \left(-\frac{3}{2},-\frac{3}{2}\right) \mathrm{with} \\ \mathrm{f}\left(-\frac{3}{2},-\frac{3}{2}\right) =-14 \end{gathered}$$