Local Max, Min, Saddle Point

Question

Find all Local Maxima, Minima and Saddle points of the function

f(x,y)=x³-3x²+3y²+y³-5

Answer 1

One thing we know about the local minima and maxima of a function of two variables is that they occur at critical points of our function. These critical points are where the gradient of our function, ∇f(x,y), satisfies, ∇f(x,y) = (f_x(x,y), f_y(x,y)) = (0,0). So every pair of points (x,y) that satisfy this equation is a critical point. It's important to note that not every critical point is a local min/max. We'll need some additional information to determine that.

A hint to make sure your heading in the right direction: In this case, you should have two values of x that satisfy f_x(x,y) = 0 and two values of y that satisfy f_y(x,y) = 0, meaning you have four critical points in total.

Second, to determine whether each point is a local max/min or saddle point, we need to implement the second derivative test. The test tells us precisely that if we let

D(x₀,y₀) = f_xx(x₀,y₀)f_yy(x₀,y₀) - (f_xy(x₀,y₀))²

for a critical point (x_0,y₀)

if D(x₀,y₀) > 0 and f_xx(x₀,y₀) >0, f(x₀,y₀) is a local min

if D(x₀,y₀) > 0 and f_xx(x₀,y₀) < 0, f(x₀,y₀) is a local max

if D(x₀,y₀) < 0, f(x₀,y₀) is a saddle point

and if D(x₀,y₀) = 0, the test is inconclusive.

For some intuition behind the second derivative test, we can think of f_xx(x₀,y₀) as the concavity of the graph in the x direction and f_yy(x₀,y₀) as the concavity in the y direction. If the second partial derivatives are both positive, then the graph must be concave up, giving us a local min. If the second partial derivatives are both negative, our first term in D is positive and the graph must be concave down in both directions, giving us a local max. However, if the partial derivatives disagree in sign, one is positive and one is negative making D <0, the graph is concave up in one direction and concave down in the other direction, not making it either a local min/max but a saddle point.

Once you determine which critical points are local maxes and mins, then you simply plug those critical points back into your function f(x,y) to get the values of the local maxes and mins.

Answer 2

f(x,y) = x³ - 3x² + 3y² + y³ - 5

∂f/∂x = 3x² - 6x = 3x(x - 2) = 0 when x = 0 or 2

∂f/∂y = 6y + 3y² = 3y(2 + y) = 0 when y = 0 or -2

∂f/∂x = 0 and ∂f/∂y = 0 at (0,0), (0,-2), (2,0), (2,-2)← critical points

D = f_xxf_yy - (f_xy)² = (6x-6)(6+6y) - (0)²

At (0,0), D = -36 < 0. So, (0,0) is a saddle point.

At (0,-2), D = 36 > 0 and f_xx = -6 < 0. So, (0,-2) is a relative max.

At (2,0), D = 36 > 0 and f_xx = 6 > 0. So, (2,0) is a relative min.

At (2,-2), D = -36 < 0 and f_xx = 6 > 0. So, (2,-2) is a relative min.