Find the maximum and minimum values of $z=f(x,y)=3x-y+1$ subject to the constraint $3x^{2}+y^{2}=9$.

Solution We use the steps for Lagrange multipliers:

Step 1 The problem is expressed in the required form.

Step 2 The functions $f$ and $g( x,y) =3x^{2}+y^{2}-9=0$ each have continuous partial derivatives.

Step 3 Solve the system of equations $$\left\{\begin{array}{rcl} 3&=&6\lambda x & \qquad {\color{#0066A7}{\hbox{\(f_{x}(x,y) =\lambda g_{x}(x,y)\)}}}\\ -1&=&2\lambda y & \qquad {\color{#0066A7}{\hbox{\(f_{y}(x,y) =\lambda g_{y}(x,y)\)}}}\\ 3x^{2}+y^{2}-9&=&0 & \qquad {\color{#0066A7}{\hspace{.3pc}\hbox{\(g(x,y) =0\)}}} \end{array}\right.$$

From the first two equations, we find that $x=\dfrac{1}{2\lambda }$ and $y=-\dfrac{1}{2\lambda }$ from which $x=-y$. Substituting into the third equation, we have $$\begin{eqnarray*} 4x^2&=&9\\[4pt] x&=&\pm \dfrac{3}{2} \end{eqnarray*}$$

Then $x=\pm \dfrac{3}{2},$ $y=\mp \dfrac{3}{2}$, and the test points are $\left( \dfrac{3}{2},-\dfrac{3}{2}\right)$ and $\left( -\dfrac{3}{2},\dfrac{3}{2}\right)$.

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Step 4 The values of $z$ corresponding to the test points are $z=\dfrac{9}{2}+\dfrac{3}{2}+1=7$ and $z=-\dfrac{9}{2}-\dfrac{3}{2}+1=-5$. The maximum value of $z$ is $7$ and the minimum value is $-5$.