Answer
The minimum value of $f\left( {x,y} \right)$ is $\frac{{36}}{{13}}$ on the constraint.
There is no maximum value of $f\left( {x,y} \right)$ on the constraint.
Work Step by Step
We have $f\left( {x,y} \right) = {x^2} + {y^2}$ and the constraint $g\left( {x,y} \right) = 2x + 3y - 6 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2x,2y} \right) = \lambda \left( {2,3} \right)$
$2x = 2\lambda $, ${\ \ \ \ }$ $2y = 3\lambda $
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$ and $y \ne 0$.
From Step 1, we obtain $\lambda = x = \frac{2}{3}y$. So $\lambda \ne 0$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2, we obtain $y = \frac{3}{2}x$. Substituting it in the constraint $g\left( {x,y} \right)$ gives
$2x + 3\left( {\frac{3}{2}x} \right) - 6 = 0$
$x = \frac{{12}}{{13}}$
There is one critical point at $\left( {\frac{{12}}{{13}},\frac{{18}}{{13}}} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical point: $\left( {\frac{{12}}{{13}},\frac{{18}}{{13}}} \right)$.
$f\left( {\frac{{12}}{{13}},\frac{{18}}{{13}}} \right) = \frac{{36}}{{13}}$
Since $f\left( {x,y} \right) = {x^2} + {y^2}$ is increasing on the constraint (please see the figure attached), we conclude that the extreme value $\frac{{36}}{{13}}$ is the minimum value of $f\left( {x,y} \right)$ on the constraint.
There is no maximum value of $f\left( {x,y} \right)$ on the constraint.
