Answer
The minimum value of $f$ subject to the constraint:
$f\left( {\frac{{18}}{7},\frac{{54}}{7},\frac{{36}}{7}} \right) = \frac{{648}}{7}$
There is no maximum value of $f$ on the constraint.
Work Step by Step
We have $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ and the constraint $g\left( {x,y,z} \right) = x + 3y + 2z - 36 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2x,2y,2z} \right) = \lambda \left( {1,3,2} \right)$
(1) ${\ \ \ \ }$ $2x = \lambda $, ${\ \ }$ $2y = 3\lambda $, ${\ \ }$ $2z = 2\lambda $
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$, $y \ne 0$, $z \ne 0$.
So, $\lambda = 2x = \frac{2}{3}y = z$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain $y=3x$ and $z=2x$. Substituting these in the constraint $g\left( {x,y,z} \right)$ gives
$x + 9x + 4x - 36 = 0$
$x = \frac{{18}}{7}$
Using $y=3x$ and $z=2x$, we obtain the critical point: $\left( {\frac{{18}}{7},\frac{{54}}{7},\frac{{36}}{7}} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical point: $f\left( {\frac{{18}}{7},\frac{{54}}{7},\frac{{36}}{7}} \right) = \frac{{648}}{7}$.
Referring to the figure attached, we notice that the surface $f$ touches the constraint $g\left( {x,y,z} \right) = 0$ at the critical point $\left( {\frac{{18}}{7},\frac{{54}}{7},\frac{{36}}{7}} \right)$. As we move $\left( {x,y,z} \right)$ along the constraint, $f$ increases. Therefore, we conclude that the critical point $\left( {\frac{{18}}{7},\frac{{54}}{7},\frac{{36}}{7}} \right)$ corresponds to the minimum value of $f$. There is no maximum value of $f$ on the constraint.
