A function of two variables has a local maximum at \((a,b)\) if \(f(x,y) \leq f(a,b)\) when \((x,y)\) is near \((a,b)\). The number \(f(a,b)\) is called a local maximum value. If \(f(x,y) \geq f(a,b)\) when \((x,y)\) is near \((a,b)\), then \(f\) has a local minimum at \((a,b)\) and \(f(a,b)\) is a local minimum value.
Fermat’s Theorem for Functions of Two Variables: If \(f\) has a local maximum or minimum at \((a,b)\) and the first-order partial derivatives of \(f\) exist there, then \(f_x(a,b) = 0\) and \(f_y(a,b) = 0\)
Suppose the second partial derivatives of \(f\) are continuous on a disk with center \((a,b)\), and suppose that \((a,b)\) is a critical point. Let
$$
\begin{aligned}
D = D(a,b) & = f_{xx}(a,b)f_{yy}(a,b) – [f_{xy}(a,b)]^2 \\
& = \left| \begin{array}{c c}
f_{xx}(a,b) & f_{xy}(a,b) \\
f_{yx}(a,b) & f_{yy}(a,b) \\
\end{array} \right|
\end{aligned}
$$
$$
\begin{aligned}
D = D(a,b) & = f_{xx}(a,b)f_{yy}(a,b) – [f_{xy}(a,b)]^2 \\
& = \left| \begin{array}{c c}
f_{xx}(a,b) & f_{xy}(a,b) \\
f_{yx}(a,b) & f_{yy}(a,b) \\
\end{array} \right|
\end{aligned}
$$
- If \(D>0\) and \(f_{xx}(a,b)>0\), then \(f(a,b)\) is a local minimum.
- If \(D>0\) and \(f_{xx}(a,b)<0\), then \(f(a,b)\) is a local maximum.
- If \(D<0\), then \(f(a,b)\) is a saddle point. That is, \(f(a,b)\) is not a local maximum or minimum.
- If \(D=0\), then the test is inconclusive.
Extreme Value Theorem for Functions of Two Variables: If \(f\) is continuous on a closed, bounded set \(D\) in \(\mathbb{R}^3\), then \(f\) attains an absolute maximum value \(f(x_1,y_1)\) and an absolute minimum value \(f(x_2, y_2)\) at some points \((x_1,y_1)\) and \((x_2, y_2)\) in \(D\).
Find the local maximum and minimums values and saddle point(s) of
$$
f(x,y) = x^2 +xy+y^2+y
$$
- Find all the critical points of \(f\), by finding solutions to the system
$$
\begin{cases}
f_x(x,y) = 2x +y = 0\\
f_y(x,y) = x + 2y +1 = 0
\end{cases}
$$
Solving the first equation for \(y\) and substituting into the second equation, the system reduces to
$$
-3x +1 = 0 \Rightarrow x = \frac{1}{3}
$$
If \(x = 1/3\) then \(y = -2/3\). Check that this solution satisfies the original system. That is, check that \(f_x(1/3,-2/3) = 0\) and \(f_y(1/3,-2/3) = 0\). - Find \(D(1/3,-2/3)\) and \(f_{xx}(1/3,-2/3)\):
$$
D(-1/3,-2/3) = \left. \left| \begin{array}{ c c}
f_{xx} & f_{xy} \\
f_{yx} & f_{yy}
\end{array} \right| \right|_{\left(\frac{1}{3},-\frac{2}{3}\right)}
$$
The second partials are
$$
\begin{aligned}
f_{xx}(x,y) & = 2 \\
f_{yy}(x,y) & = 2 \\
f_{xy}(x,y) & = 1
\end{aligned}
$$
So that
$$
D(-1/3,-2/3) = 4-1 = 3 >0
$$
Since \(D(1/3,-2/3)>0\) and \(f_{xx}(1/3,-2/3)>0\), the point \((1/3,-2/3)\) is a local minimum with minimum value \(f(1/3,-2/3) =-1/3 \)
Find the local maximum and minimums values and saddle point(s) of
$$
f(x,y) = (x^2+y^2)e^{y^2-x^2}
$$
- Find all the critical points of \(f\), by finding solutions to the system
$$
\begin{cases}
f_x(x,y) = -2 x e^{y^2-x^2} \left(x^2+y^2-1\right) = 0\\
f_y(x,y) = 2 y e^{y^2-x^2} \left(x^2+y^2+1\right) = 0
\end{cases}
$$
Notice that the second equation is satisfied if and only if \(y = 0\). Substituting this value of \(y\) into the first equation reduces the system to
$$
-2xe^{y^2-x^2}(x^2-1) = 0
$$
That is, \(x = 0\) or \( x= \pm 1\). So there are three critical points:- \((0,0)\)
- \((1,0)\)
- \((-1,0)\)
- Find \(D(a,b)\) and \(f_{xx}(a,b)\) for each of the critical points:
The second partials are
$$
\begin{aligned}
f_{xx}(x,y) & = 2 e^{y^2-x^2} \left(2 x^4+2 x^2 y^2-5 x^2-y^2+1\right) \\
f_{yy}(x,y) & = 2 e^{y^2-x^2} \left(2 x^2 y^2+x^2+2 y^4+5 y^2+1\right) \\
f_{xy}(x,y) & = -4 x y e^{y^2-x^2} \left(x^2+y^2\right)
\end{aligned}
$$- \(D(0,0) = 4\) and \(f_{xx}(0,0) = 2 \)
- \(D(1,0) = -\frac{16}{e^2}\) and \(f_{xx}(1,0) = -\frac{4}{e} \)
- \(D(-1,0) = -\frac{16}{e^2} \) and \(f_{xx}(-1,0) = -\frac{4}{e}\)
Since \(D(0,0)>0\) and \(f_{xx}(0,0)>0\), the point \((0,0)\) is a local minimum with minimum value \(f(0,0) =0 \). The other critical points are saddle points since \(D<0\) for these points.