16. Max-Min Problems

a. Local Minima, Local Maxima and Saddle Points

3. Second Derivative Test - Examples

Find all critical points of the function \(f(x,y)=5x^2-2x^3-xy^2+y^2\). Use the second derivative test to identify each as a local maxima, local minima or saddle or say the test fails.

To find the critical points, we compute the partial derivatives and set them equal to zero. \[\begin{aligned} f_x&=10x-6x^2-y^2=0 \\ f_y&=-2xy+2y=0 \end{aligned}\] We need to solve the two equations for the two unknowns, \(x\) and \(y\). They cannot be solved separately. They must be solved simultaneously by looking at cases. This is called case work. Since the \(f_y\) equation factors, we start with that equation: \[ -2y(x-1)=0 \] This tells us that either \(y=0\) or \(x=1\). These are the two cases:

• Case 1: \(y=0\): The \(f_x\) equation reduces to \(10x-6x^2=0\) which factors as \(2x(5-3x)=0\) and has the solutions \(x=0\) and \(x=\dfrac{5}{3}\). This leads to the two critical points \((0,0)\) and \(\left(\dfrac{5}{3},0\right)\).
• Case 2: \(x=1\): The \(f_x\) equation reduces to \(10-6-y^2=0\) which factors as \((y-2)(y+2)=0\) and has the solutions \(y=2\) and \(y=-2\). This leads to the two critical points \((1,2)\) and \((1,-2)\).

Now that we have the four critical points, we use the second derivative test to classify each as a maximum, minimum, or saddle point. First, the second derivatives are: \[\begin{aligned} f_{xx}&=10-12x \\ f_{yy}&=-2x+2 \\ f_{xy}&=-2y \end{aligned}\] We make a table to keep track of all the values. There is no need to compute a separate formula for \(D=f_{xx}f_{yy}-{f_{xy}}^2\). We just do it as needed in the table.

So \((0,0)\) is a local minimum, \(\left(\dfrac{5}{3},0\right)\) is a local maximum, while \((1,2)\) and \((1,-2)\) are saddle points.