Before we work on maxima and minima for 3D functions, it might help to review what we know about 2D functions. We found critical points when the first derivative of a function is zero — when the tangent line to the function is horizontal:
$$\frac{df}{dx} = 0$$
For example, recall that the parabola $f(x) = x^2$ has a minimum at $(0, 0)$, and that
$$\frac{df}{dx} = 2x$$
is zero at $x = 0$, where $y = 0$. And we can confirm that $(0,0)$ is a minimum by the second-derivative test:
$$\frac{d^2 f}{dx^2} = 2 \gt 0.$$
Because the second derivative is positive, the function is concave upward (everywhere in this case). Here's the function with its derivative at $(0,0)$.
Likewise for $f(x) = -x^2$, which we know to be an upside-down parabola,
$$\frac{df}{dx} = -2x \phantom{00} \text{ and } \phantom{00} \frac{d^2 f}{dx^2} = -2$$
which gives us a critical point at $(0,0)$ and a negative second derivative, which means concave-downward. Here's the graph:
We call these local maxima and minima, but in these cases they are the global maximum and minimum, respectively. Now let's look at one more 2D function, $f(x) = x^3$. The critical point is
$$\frac{df}{dx} = 3x^2 = 0 \; \rightarrow \; x = 0$$
The second derivative is $6x$, and evaluated at the critical point $(0, 0)$ we get $0$, which is indeterminate for the curvature of the function — it's between concave upward and concave downward. For 2D graphs, we know this kind of point as an inflection point, a point where the curvature changes between the two concavities. Here's the graph:
For functions in $\mathbb{R}^3$ — functions of two variables like $f(x,y) = x^2 - y^2$, shown below, we'll be able to make some analogies with the 2D case.
To have a critical point in a function $f(x,y)$, we require that both partial derivatives equal zero:
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$$
We can find those partials for $f(x,y) = x^2 - y^2$ and set them equal to zero to find our critical point(s):
$$\frac{\partial f}{\partial x} = 2x \phantom{00} \text{and} \phantom{00} \frac{\partial f}{\partial y} = 2y$$
$$2x = 0 \; \rightarrow x = 0 \phantom{00} \text{and} \phantom{00} 2y = 0 \; \rightarrow y = 0$$
Our single critical point is $(x, y) = (0, 0)$.
Now our partial derivative with respect to $x$ treats $y$ as a constant, and locks us into the $x-z$ plane. That leaves us with the parabola $z = x^2$ in
$$\frac{\partial^2 f}{\partial x^2} = 2 \gt 0$$
... so that matches our finding from the 2D case. Likewise, the green, downward-opening parabola traced in the $y-z$ plane has a negative second partial derivative,
$$\frac{\partial^2 f}{\partial y^2} = -2 \lt 0$$
So our critical point is both a local minimum (in the $xz$ plane) and a local maximum (in the $yz$ plane). This kind of point, which is analogous to the inflection point in 2D, is called a saddle point.
Now we're going to jump straight to the general second-derivative test for multidimensional functions without much explanation. Then we'll rationalize it afterward. Here are the key elements of the 2^{nd} derivative test:
Calculate the 2D determinant,
$$ \begin{align} D &= \text{det} \left( \begin{array}{rr} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{array} \right) \\[5pt] &= f_{xx} f_{yy} - (f_{xy})^2 \end{align}$$
If $D(a,b) \gt 0$ and $f_{xx}(a,b) \gt 0$, then $(a, b)$ is a local minimum of $f(x,y)$.
If $D(a,b) \gt 0$ and $f_{xx}(a,b) \lt 0$, then $(a, b)$ is a local maximum of $f(x,y)$.
If $D(a,b) \lt 0$, then $(a, b)$ is a saddle point of $f(x,y)$.
Let's perform this second-derivative test for our function $f(x,y)=x^2-y^2$. We have
$$ \begin{align} f_{xx} &= 2 \phantom{00} f_{yy} = -2 \phantom{00} f_{xy} = 0 \\[10pt] D &= f_{xx} f_{yy} - (f_{xx})^2 \\[5pt] &= 2(-2) - 0 = -4 \lt 0 \end{align}$$
Now look at our value $D$. If $f_{xx}$ and $f_{yy}$ are of opposite signs, as they are in our example, then $f_{xx}f_{yy}$ is negative, so $D = f_{xx} f_{yy} - (f_{xx})^2$ must be negative, and we have a saddle point.
If both $f_{xx}$ and $f_{yy}$ have the same sign, then $D$ can be positive or negative. In the latter case, $f_{xx}$ evaluated at the critical point, is positive, then so is $f_{yy}$, and curves analogous to our two parabolas are both concave-upward, so we have a local minimum. Conversely, if $f_{xx} \lt 0$ then we have a local maximum.
If $D = 0$ the second derivative test is inconclusive.
Find and characterize any critical points of the function $f(x,y) = x \, e^{-2x^2-2y^2}$
First calculate the first partial derivatives — this is a tricky function requiring the product and chain rules; be careful. I've skipped some steps for brevity in calculating these partial derivatives.
$$ \begin{align} f_x &= e^{-2x^2-2y^2}[1-4x^2] = 0 \\[5pt] 1 - 4x^2 &= 0 \tag{1} \\[5pt] x &= \pm \frac{1}{2} \\[8pt] f_y &= -4xy \, e^{-2x^2-2y^2} = 0 \\[5pt] y &= 0 \end{align}$$
Notice that in (1), we used the fact that $e^{-2x^2-2y^2}$ is never equal to zero, which makes finding the zeros of the first partials. So our critical points are $\left( \frac{1}{2}, 0 \right), \; \left( -\frac{1}{2}, 0 \right)$.
Now our second derivatives are
$$ \begin{align} f_{xx} &= e^{-2x^2-2y^2}[16x^3 - 12x] \\[5pt] f_{yy} &= e^{-2x^2-2y^2}[-4x+16y^2] \\[5pt] f_{xy} &= -4y \, e^{-2x^2-2y^2}[1-4x^2] \end{align}$$
As a shorcut, we can calculate $e^{-2x^2-2y^2}$ for $x = \pm \frac{1}{2}, \; y = 0$. It's value (regardless of the sign of $x$) is $e^{1/2}$. What will matter most to us is that it's a positive number.
Now we can calculate
$$ \begin{align} f_{xx}\left( \frac{1}{2}, 0 \right) &= \frac{-4}{\sqrt{e}} \\[5pt] f_{xx}\left( -\frac{1}{2}, 0 \right) &= \frac{4}{\sqrt{e}} \\[5pt] f_{yy}\left(\pm \frac{1}{2}, 0 \right) &= \frac{-2}{\sqrt{e}} \\[5pt] f_{xy} \left(\pm \frac{1}{2}, 0 \right) &= 0 \end{align}$$
Now we can caclulate the determinant, $D$
$$D = \text{det} \left( \begin{array}{rr} \frac{-4}{\sqrt{e}} & 0 \\ 0 & \frac{-2}{\sqrt{e}} \end{array} \right) = \frac{8}{e} \gt 0 $$
Finally, we can evaluate each critical point:
$$ \begin{align} \left( \frac{1}{2}, 0 \right): &\phantom{00} D \gt 0 \phantom{00} f_{xx} \lt 0 \rightarrow \text{max} \\[5pt] \left( -\frac{1}{2}, 0 \right): &\phantom{00} D \gt 0 \phantom{00} f_{xx} \gt 0 \rightarrow \text{max} \end{align}$$
Our critical points include a local maximum (which is also the global max in this case) at $(x,y) = \left( \frac{1}{2}, 0 \right)$ and a local minimum (also the global min) at $(x,y) = \left( -\frac{1}{2}, 0 \right)$. These findings are consistent with the 3D plot below:
This function does not have a saddle point.
Find and characterize any critical points of the function $f(x,y) = x - x^2y - y + xy^2$
The first partial derivatives are
$$ \begin{align} f_x &= 1 - 2xy + y^2 = 0\\[5pt] f_y &= -1 + 2xy - x^2 = 0 \end{align}$$
If we add those two equations, we can eliminate the first two terms of each:
$$y^2 - x^2 = 0 \; \rightarrow \; x = \pm y$$
Now let's plug that result into $f_x$, including the $\pm$.
$$ \begin{align} f_x = 1 - 2x(\pm x) + (\pm x)^2 &= 0 \\[5pt] 1 \pm 2x^2 + x^2 &= 0 \\[5pt] \end{align}$$
This reduces to $1 - x^2 = 0$ for the (+) solution, which gives $x = \pm 1$. For the (-) solution there are no real zeros, so we have $x = \pm 1$. Now we plug that in (with the $\pm$) to either $f_x$ or $f_y$:
$$ \begin{align} -1 + 2(\pm 1) y - (\pm y)^2 &= 0 \\[5pt] -1 \pm 2y - 1 &= 0 \\[5pt] \pm 2y &= 2 \\[5pt] y &= \pm 1 \end{align}$$
So our candidates for critical points are $(1, 1), \; (1, -1), \; (-1, 1)$ and $(-1, -1)$.
Now our second-derivative test. The second derivatives are
$$f_{xx} = -2y \phantom{000} f_{yy} = 2x\phantom{000} f_{xy} = -2(x-y)$$
Then our determinant is
$$D = \text{det} \left( \begin{matrix} -2y & -2(x-y) \\ -2(x-y) & 2y \end{matrix} \right) = -4y^2 -4(x-y)^2$$
Here is a table of our $D$ and $f_{xx}$ values at our four candidate critical points.
Qty. | (1,1) | (1,-1) | (-1,1) | (-1,-1) |
---|---|---|---|---|
$D(a,b)$ | -4 | -20 | -20 | -4 |
$f_{xx}(a,b)$ | -2 | 2 | -2 | 2 |
$f_{yy}(a,b)$ | 2 | 2 | -2 | -2 |
All of the $D$ values are negative, indicating saddle points rather than local maxima or minima. But in this case we need to look closer — that's why both second partials $f_{xx}$ and $f_{yy}$ are included in the table. For a true saddle point, the partial derivatives in the $x$ and $y$ directions must have opposite signs. that is only true for our critical points $(1, 1)$ and $(-1,-1)$.
Here's a 3-D plot showing those saddle points. Notice that this function has no local maxima or minima.
Find and characterize any critical points of the function $f(x,y) = (x^2+y^2)e^{-x}$
The first partial derivatives are
$$ \begin{align} f_x &= -e^{-x}(x^2 + y^2) + e^{-x}(2x) \\[5pt] &= e^{-x}[2x - (x^2 + y^2)] \\[5pt] &= e^{-x}[-x^2 - y^2 + 2x] = 0\\[8pt] f_y &= -2y e^{-x} = 0 \\[5pt] &\rightarrow y = 0 \end{align}$$
Now plug $y = 0$ into the $f_x$ expression:
$$ \begin{align} f_x(y = 0) &= -e^{-x}[-x^2 - 0 + 2x] \\[5pt] x^2 &= 2x \\[5pt] &\rightarrow \; x = 0 \text{ or } 2 \end{align}$$
So our critical points are $(0, 0)$ and $(2, 0)$. Now the second partials:
$$f_{xx} = e^{-x}[x^2 + y^2 - 4x + 2] \phantom{000} f_{yy} = 2 e^{-x}$$
$$f_{xy} = f{yx} = -2y \, e^{-x}$$
The discriminant, $D$, is
$$D = \text{det} \left( \begin{matrix} e^{-x}[-x^2-y^2+2x] & -2y \, e^{-x} \\ -2y \, e^{-x} & 2y \, e^{-x} \end{matrix} \right) = 2e^{-2x}[-x -y^2 + 2x + 2y]$$
Now we can make a table of our second-derivative test key values:
Qty. | (0,0) | (2,0) |
$D(a,b)$ | 4 | -4 e^{-4} |
$f_{xx}(a,b)$ | 2 | 2 e^{-2} |
Finally, critical point $(0, 0)$ is a local minimum because $D \gt 0$ and $f_{xx} \gt 0$ (the curvature is concave upward). And $(2, 0)$ is a saddle point because $D \lt 0$. Here's a graph of that function.
Find the critical points of the following $\mathbb{R}^3$ functions, and determine whether each is a local minimum, a local maximum or a saddle point.
$f(x,y) = x^2-y-ln(x+y)$
$$ \begin{align} f_x &= 2x - \frac{1}{x+y} = 0 \\[5pt] 2x &= \frac{1}{x+y} \\[5pt] 2x^2 + 2xy - 1 &= 0 \tag{1} \\[8pt] f_y &= -1 - \frac{1}{x+y} \\[5pt] y &= -x - 1 \tag{2} \end{align}$$
We can solve for $x$ by inserting (2) into (1):
$$ \begin{align} 2x^2 + 2x(-x-1) - 1 &= 0 \\[5pt] -2x &= 1 \\[5pt] x &= -\frac{1}{2} \end{align}$$
Then we use this in $y = -x -1$ to get $y = -\frac{1}{2}$, thus our one critical point is $\left( -\frac{1}{2}, -\frac{1}{2} \right)$.
We can actually stop here because that point is not in the domain of our function — we can't take the log of a negative number. So we conclude that this function has no local minima, maxima or saddle points. Here's a graph:
$f(x,y) = ln(x) + 4ln(y) - x - 4y$
$$ \begin{align} f_x &= \frac{1}{x}-1 = 0 \\[5pt] x &= 1 \\[5pt] f_y &= \frac{4}{y} - 4 = 0 \\[5pt] y &= 1 \end{align}$$
So our only critical point is $(1, 1)$.
Then we use this in $y = -x -1$ to get $y = -\frac{3}{2}$, thus our one critical point is $\left( -\frac{1}{2}, -\frac{3}{2} \right)$.
The second partials are
$$ \begin{align} f_{xx} &= \frac{-1}{x^2} \\[5pt] f_{yy} &= \frac{-4}{x^2} \\[5pt] f_{xy} &= 0 \end{align}$$
$$D = \text{det} \left( \begin{matrix} \frac{-1}{x^2} & 0 \\ 0 & \frac{-4}{y^2} \end{matrix} \right) = \frac{4}{x^2 y^2}$$
$D$ is always positive. Then $f_{xx}(1,1) = -1$. Because this is less than zero and $D > 0$, the critical point is a local maximum.
$f(x,y) = x^3 + 2xy - 2y^2 - 10x$
$$ \begin{align} f_x &= 3x^2+2y-10 = 0 \\[5pt] f_y &= 2x - 4y = 0 \\[5pt] 4y &= 2x \; \rightarrow \; x = 2y \end{align}$$
Plugging $x=2y$ into the $f_x$ equation gives
$$ \begin{align} 3(2y)^2 + 2y -10 &= 0 \\[5pt] 4(4y^2) + 2y -10 &= 0 \\[5pt] 12y^2 + 2y &= 10 \\[5pt] y^2 + \frac{y}{6} + \left( \frac{1}{12} \right)^2 &= \frac{5}{6} + \frac{1}{144} \tag{*} \\[5pt] \left( y + \frac{1}{12} \right)^2 &= \frac{120 + 1}{144} \tag{*} \\[5pt] y &= \frac{-1 \pm \sqrt{121}}{\sqrt{144}} \tag{*} \\ y &= \frac{-1 \pm 11}{12} = -1, \frac{5}{6} \end{align}$$
The starred steps above are solving the quadratic by completing the square. We can get the x-coordinates for each of those $y$ values using $f_y = 0$:
$$ \begin{align} f_y(y = -1) = 2x + 4 = 0 \; \rightarrow \; x &= -2 \\[5pt] f_y\left(y = \frac{5}{6} \right) = 2x - 4 \left( \frac{5}{6} \right) = 0 \; \rightarrow \; x &= \frac{5}{3} \end{align}$$
So our two critical points are $(-2, -1)$ and $\left( \frac{5}{3}, \frac{5}{6} \right)$
The second partials are
$$ \begin{align} f_{xx} &= 6x \\[5pt] f_{yy} &= -4 \\[5pt] f_{xy} &= 2 \end{align}$$
$$D = \text{det} \left( \begin{matrix} 6x & 2 \\ 2 & -4 \end{matrix} \right) = \frac{4}{x^2 y^2} = -24x - 4$$
Here is a table of $D$ and $f_{xx}$ values for our two critical points.
Qty. | (-2,-1) | (5/3,5/6) |
$D(a,b)$ | 44 | -24 |
$f_{xx}(a,b)$ | -12 | 10 |
Based on these quantities, $(-2,-1)$ is a local maximum and $\left( \frac{5}{3}, \frac{5}{6} \right)$ is a saddle point.
$f(x,y) = x^2 + y^2 - xy + x$
The first partial derivatives are
$$ \begin{align} f_x &= 2x - y + 1 = 0 \\[5pt] f_y &= 2y -x = 0 \; \rightarrow \; y = \frac{x}{2} \\[5pt] \end{align}$$
Plugging that value of y back into $f_x$ gives
$$ \begin{align} 2x - \frac{x}{2} + 1 &= 0 \\[5pt] \frac{3}{2} y = -1 \; \rightarrow \; y &= -\frac{3}{2} \end{align}$$
The $x$-coordinate is $-\frac{2}{3}$, so our critical point is $\left( -\frac{2}{3}, -\frac{3}{2} \right)$.
The partial derivatives are
$$ \begin{align} f_{xx} = 2 \\[5pt] f_{yy} = 2 \\[5pt] f_{xy} = -1 \end{align}$$
That makes $D$ easy, $D = 3 \gt 0$. Now $f_{xx}(-2/3,-1/3) = 2$, also positive, so the critical point must be a local minimum.
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