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Multivariable Optimization using The Second Derivative Test (Example 1)
We can use a tool called the “second derivative test” to classify extreme points in a multivariate function.
Before, calculus with one variable just involved finding the first and second derivative of the function.
Now, we need to find all the first partial derivatives and also all second partial derivatives to classify the extreme points.
Find all the local maximums, minimums, and saddle points for \(f(x,y) = {y^3} + 3{x^2}y – 6{x^2} – 6{y^2} + 2\)
The first step is to find the first partial derivatives, set each to zero, and find as many of the extreme points as possible.
I will use f with subscripts to denote the partial derivatives here.
\[{f_x} = 6xy – 12x\]
\[6xy – 12x = 0\]
From this, x = 0 or y = 2.
\[{f_y} = 3{x^2} + 3{y^2} – 12y\]
\[{f_y} = {x^2} + {y^2} – 4y\]
\[{x^2} + {y^2} – 4y = 0\]
\[{y^2} – 4y = – {x^2}\]
We will need to substitute x = 0 and y = 2 into this equation to find the extreme points.
First let’s do x = 0:
\[{y^2} – 4y = 0\]
So when x = 0, y = 0 and y = 4.
Now do this for y = 2 and solve for x:
\[{x^2} + {(2)^2} – 4(2) = 0\]
\[{x^2} = 4\]
So when y = 2, x = 2 and x = -2.
So all the extreme points are (0,0), (0,4), (2,2), and (-2,2).
When working with a complicated function and asked this on an exam, be careful and look for as many solutions as possible.
I’ve seen students often forget an extreme point entirely, resulting in many lost marks.
The next step is to find all of the second derivatives.
Note that \({f_{xy}} = {f_{yx}}\) at our scope of study, which just means if you differentiate x then y is the same if you differentiated y then x.
By chance xx and yy derivatives ended up being the same, usually that’s not the case!
\[{f_{xx}} = 6y – 12\]
\[{f_{yy}} = 6y – 12\]
\[{f_{xy}} = 6x\]
The second derivative test is (you may need to memorize it or ask your teacher/professor if it’s on a cheat sheet):
\[D = {f_{xx}} \cdot {f_{yy}} – {\left[ {{f_{xy}}} \right]^2}\]
And you check the sign of D for each possible point.
If D > 0 and \({f_{xx}}\) < 0, the point is a local maximum.
If D > 0 and \({f_{xx}}\) > 0, the point is a local minimum.
If D < 0, the point is a saddle point.
If D = 0, the test is inconclusive!
Let’s try the second derivative for the point (0,4) for example:
\[D(0,4) = (12)(12) – {(0)^2}\]
\[D(0,4) = 144\]
D is positive (144) and fxx is positive (12) so the point (0,4) is a local minimum.
Repeating for the point (0,0), we get D = 144 and fxx < 0, so (0,0) is a local maximum.
Checking (2,2) and (-2,2) result in D = -144, meaning that those two points are saddle points.
Thus we’ve analyzed the multivariate function’s extreme points using the second derivative test.
The main difficulty in these problems is making sure you solved for all the extreme points and remembering how to do partial derivatives.