Find the maximum and minimum values of \(z=f(x,y)=3x-y+1\) subject to the constraint \(3x^{2}+y^{2}=9\).
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 \(-5\).