Second partial derivative test

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In mathematics, the second partial derivatives test is a method in multivariable calculus used to determine if a critical point (x, y) is a minimum, maximum or saddle point.

Suppose that

M = f_{xx}(a,b)f_{yy}(a,b) - \left( f_{xy}(a,b) \right)^2

or in other words the determinant of a 2×2 Hessian matrix.

If M > 0 and fxx(a,b) > 0 then f(a, b) is a local minimum.

If M > 0 and fxx(a,b) < 0 then f(a, b) is a local maximum.

If M < 0 then f(a, b) has a saddle point.

If M = 0 then the second derivatives test is indecisive.

We only test values for which fx = 0 and fy = 0. This is rationalized because the function must on its trace along the xz-plane have its derivative equal to zero and the same is true for a trace on the yz-plane.

Find and label the critical points of the following function:

z = (x + y)(xy + xy2)

To solve this problem we must first find the first partial derivatives with respect to x and y of the function.

\frac{\partial z}{\partial x} = y(2x +y)(y+1)
\frac{\partial z}{\partial y} = x \left( 3y^2 +2y(x+1) + x \right)

Looking at

\frac{\partial z}{\partial x} \left( x \right)

So x must equal zero for y = 0.

we see that y must equal 0, −1 or − 2x

We plug this into the next equation, we get

\frac{\partial z}{\partial y} = x \left( 3y^2 +2y(x+1) + x \right) = x

There were other possibilities for y, so we have

\frac{\partial z}{\partial y} = x \left( 3 -2(x+1) + x \right) = 0 = x(1-x)

So x must be equal to 1 or 0.

\frac{\partial z}{\partial y} = x \left( 3(-2x)^2 +2(-2x)(x+1) + x \right) = 4x^2(2x-1)
So x must equal 0 or \frac{1}{2}

Let's list all the critical values now.

(x,y) \in {(0,0), (0, -1), (1,-1), (\frac{1}{2}, -1)}

Now we have to label the critical values using the second derivative test.

D = f_{xx}(a,b)f_{yy}(a,b) - \left( f_{xy}(a,b) \right)^2 = 2y(y+1)(2(3y+x+1))x - (3y^2+y(4x+2)+2x)^2

Now we plug in all the different critical values we found to label them.

At (0, 0) we have D = 0, at (0, −1); D = −1, at (1, −1); D = −1, at

(\frac{1}{2}, -1) D = 0.

So we can now label some of the points, at (0, −1) and (1, −1) f(x, y) has a saddle point. At the other two points we need higher order tests to find out what exactly the function is doing.

Application of the second partial derivatives test is fairly straightforward and it is usually used as an all-purpose tool for identifying what the critical values of a function are, i.e. what the point (a, b) is on f(x, y). However we must not simply attack every problem of maximization with this or the method of Lagrange multipliers, as more trivial methods may prove to be both simpler and more elegant, Consider the following example:

Given the paraboloid

f(x,y) = x2 + y2

and we are asked to find and identify the critical values of the function, we can immediately conclude that the origin is the absolute minimum, since all the level curves are circles of increasing radius.

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