4.8 Lagrange multipliers  (Page 4/11)

Let’s follow the problem-solving strategy:

1. The optimization function is $f\left(x,y,z\right)=x^2+y^2+z^2.$ To determine the constraint functions, we first subtract $z^2$ from both sides of the first constraint, which gives $x^2+y^2-z^2=0,$ so $g\left(x,y,z\right)=x^2+y^2-z^2.$ The second constraint function is $h\left(x,y,z\right)=x+y-z+1.$
2. We then calculate the gradients of $f,g,\text{and}h\text{:}$
   
   $\begin{array}{c}\nabla f\left(x,y,z\right)=2xi+2yj+2zk\hfill \\ \nabla g\left(x,y,z\right)=2xi+2yj-2zk\hfill \\ \nabla h\left(x,y,z\right)=i+j-k.\hfill \end{array}$
   
   The equation $\nabla f\left(x_0,y_0,z_0\right)={\lambda }_1\nabla g\left(x_0,y_0,z_0\right)+{\lambda }_2\nabla h\left(x_0,y_0,z_0\right)$ becomes
   
   $2x_{0}i+2y_{0}j+2z_{0}k={\lambda }_1\left(2x_{0}i+2y_{0}j-2z_{0}k\right)+{\lambda }_2\left(i+j-k\right),$
   
   which can be rewritten as
   
   $2x_{0}i+2y_{0}j+2z_{0}k=\left(2{\lambda }_1{x}_0+{\lambda }_2\right)i+\left(2{\lambda }_1{y}_0+{\lambda }_2\right)j-\left(2{\lambda }_1{z}_0+{\lambda }_2\right)k.$
   
   Next, we set the coefficients of $i\text{and}j$ equal to each other:
   
   $\begin{array}{c}2x_0=2{\lambda }_1{x}_0+{\lambda }_2\hfill \\ 2y_0=2{\lambda }_1{y}_0+{\lambda }_2\hfill \\ 2z_0=\mathrm{-2}{\lambda }_1{z}_0-{\lambda }_2.\hfill \end{array}$
   
   The two equations that arise from the constraints are $z_0{}^2=x_0{}^2+y_0{}^2$ and $x_0+y_0-z_0+1=0.$ Combining these equations with the previous three equations gives
   
   $\begin{array}{ccc}\hfill 2x_0& =\hfill & 2{\lambda }_1{x}_0+{\lambda }_2\hfill \\ \hfill 2y_0& =\hfill & 2{\lambda }_1{y}_0+{\lambda }_2\hfill \\ \hfill 2z_0& =\hfill & \mathrm{-2}{\lambda }_1{z}_0-{\lambda }_2\hfill \\ \hfill z_0{}^2& =\hfill & x_0{}^2+y_0{}^2\hfill \\ \hfill x_0+y_0-z_0+1& =\hfill & 0.\hfill \end{array}$
3. The first three equations contain the variable ${\lambda }_2.$ Solving the third equation for ${\lambda }_2$ and replacing into the first and second equations reduces the number of equations to four:
   
   $\begin{array}{ccc}\hfill 2x_0& =\hfill & 2{\lambda }_1{x}_0-2{\lambda }_1{z}_0-2z_0\hfill \\ \hfill 2y_0& =\hfill & 2{\lambda }_1{y}_0-2{\lambda }_1{z}_0-2z_0\hfill \\ \hfill z_0{}^2& =\hfill & x_0{}^2+y_0{}^2\hfill \\ \hfill x_0+y_0-z_0+1& =\hfill & 0.\hfill \end{array}$
   
   Next, we solve the first and second equation for ${\lambda }_1.$ The first equation gives ${\lambda }_1=\frac{x_0+z_0}{x_0-z_0},$ the second equation gives ${\lambda }_1=\frac{y_0+z_0}{y_0-z_0}.$ We set the right-hand side of each equation equal to each other and cross-multiply:
   
   $\begin{array}{ccc}\hfill \frac{x_0+z_0}{x_0-z_0}& =\hfill & \frac{y_0+z_0}{y_0-z_0}\hfill \\ \hfill \left(x_0+z_0\right)\left(y_0-z_0\right)& =\hfill & \left(x_0-z_0\right)\left(y_0+z_0\right)\hfill \\ \hfill x_0{y}_0-x_0{z}_0+y_0{z}_0-z_0{}^2& =\hfill & x_0{y}_0+x_0{z}_0-y_0{z}_0-z_{0}2\hfill \\ \hfill 2y_0{z}_0-2x_0{z}_0& =\hfill & 0\hfill \\ \hfill 2z_0\left(y_0-x_0\right)& =\hfill & 0.\hfill \end{array}.$
   
   Therefore, either $z_0=0$ or $y_0=x_0.$ If $z_0=0,$ then the first constraint becomes $0=x_0{}^2+y_0{}^2.$ The only real solution to this equation is $x_0=0$ and $y_0=0,$ which gives the ordered triple $\left(0,0,0\right).$ This point does not satisfy the second constraint, so it is not a solution.
   Next, we consider $y_0=x_0,$ which reduces the number of equations to three:
   
   $\begin{array}{ccc}\hfill y_0& =\hfill & x_0\hfill \\ \hfill z_0{}^2& =\hfill & x_0{}^2+y_0{}^2\hfill \\ \hfill x_0+y_0-z_0+1& =\hfill & 0.\hfill \end{array}$
   
   We substitute the first equation into the second and third equations:
   
   $\begin{array}{ccc}\hfill z_0{}^2& =\hfill & x_0{}^2+x_0{}^2\hfill \\ \hfill x_0+x_0-z_0+1& =\hfill & 0.\hfill \end{array}$
   
   Then, we solve the second equation for $z_0,$ which gives $z_0=2x_0+1.$ We then substitute this into the first equation,
   
   $\begin{array}{ccc}\hfill z_0{}^2& =\hfill & 2x_0{}^2\hfill \\ \hfill {\left(2x_0+1\right)}^2& =\hfill & 2x_0{}^2\hfill \\ \hfill 4x_0{}^2+4x_0+1& =\hfill & 2x_0{}^2\hfill \\ \hfill 2x_0{}^2+4x_0+1& =\hfill & 0,\hfill \end{array}$
   
   and use the quadratic formula to solve for $x_0\text{:}$
   
   $x_0=\frac{\mathrm{-4}\pm \sqrt{4^2-4\left(2\right)\left(1\right)}}{2\left(2\right)}=\frac{\mathrm{-4}\pm \sqrt{8}}{4}=\frac{\mathrm{-4}\pm 2\sqrt{2}}{4}=\mathrm{-1}\pm \frac{\sqrt{2}}{2}.$
   
   Recall $y_0=x_0,$ so this solves for $y_0$ as well. Then, $z_0=2x_0+1,$ so
   
   $z_0=2x_0+1=2\left(\mathrm{-1}\pm \frac{\sqrt{2}}{2}\right)+1=\mathrm{-2}+1\pm \sqrt{2}=\mathrm{-1}\pm \sqrt{2}.$
   
   Therefore, there are two ordered triplet solutions:
   
   $\left(\mathrm{-1}+\frac{\sqrt{2}}{2},\mathrm{-1}+\frac{\sqrt{2}}{2},\mathrm{-1}+\sqrt{2}\right)\text{and}\left(\mathrm{-1}-\frac{\sqrt{2}}{2},\mathrm{-1}-\frac{\sqrt{2}}{2},\mathrm{-1}-\sqrt{2}\right).$
4. We substitute $\left(\mathrm{-1}+\frac{\sqrt{2}}{2},\mathrm{-1}+\frac{\sqrt{2}}{2},\mathrm{-1}+\sqrt{2}\right)$ into $f\left(x,y,z\right)=x^2+y^2+z^2,$ which gives
   
   $\begin{array}{cc}\hfill f\left(\mathrm{-1}+\frac{\sqrt{2}}{2},\mathrm{-1}+\frac{\sqrt{2}}{2},\mathrm{-1}+\sqrt{2}\right)& ={\left(\mathrm{-1}+\frac{\sqrt{2}}{2}\right)}^2+{\left(\mathrm{-1}+\frac{\sqrt{2}}{2}\right)}^2+{\left(\mathrm{-1}+\sqrt{2}\right)}^2\hfill \\ & =\left(1-\sqrt{2}+\frac{1}{2}\right)+\left(1-\sqrt{2}+\frac{1}{2}\right)+\left(1-2\sqrt{2}+2\right)\hfill \\ & =6-4\sqrt{2}.\hfill \end{array}$
   
   Then, we substitute $\left(\mathrm{-1}-\frac{\sqrt{2}}{2},\mathrm{-1}-\frac{\sqrt{2}}{2},\mathrm{-1}-\sqrt{2}\right)$ into $f\left(x,y,z\right)=x^2+y^2+z^2,$ which gives
   
   $\begin{array}{cc}\hfill f\left(\mathrm{-1}-\frac{\sqrt{2}}{2},\mathrm{-1}-\frac{\sqrt{2}}{2},\mathrm{-1}-\sqrt{2}\right)& ={\left(\mathrm{-1}-\frac{\sqrt{2}}{2}\right)}^2+{\left(\mathrm{-1}-\frac{\sqrt{2}}{2}\right)}^2+{\left(\mathrm{-1}-\sqrt{2}\right)}^2\hfill \\ & =\left(1+\sqrt{2}+\frac{1}{2}\right)+\left(1+\sqrt{2}+\frac{1}{2}\right)+\left(1+2\sqrt{2}+2\right)\hfill \\ & =6+4\sqrt{2}.\hfill \end{array}$
   
   $6+4\sqrt{2}$ is the maximum value and $6-4\sqrt{2}$ is the minimum value of $f\left(x,y,z\right),$ subject to the given constraints.