Extrema examples

Example 1

Find the local extrema of f (x, y) = x³ + x²y - y² - 4y.

Solution

Step 1: Find the critical points. The derivative of f is

$$\left[\vphantom{3x^2+2xy \quad x^2-2y-4}\right. \left.\vphantom{3x^2+2xy \quad x^2-2y-4}\right]$$

Df (x, y) = [0    0] means both components must be zero simultaneously. We need

x(3x + 2y) = 0 (1)

and

x² - 2y - 4 = 0. (2)

We need to solve two equations for the two unknowns x and y.

Equation (1) is satified if either x = 0 or if 3x + 2y = 0, i.e., if x = 0 or if y = - 3x/2. We consider these two solutions as two separate cases. For each case, we will find soultions for equation (2).

Case 1: Let x = 0. Then we know equation (1) is satisfied. We plug x = 0 into equation (2), which becomes 0 - 2y - 4 = 0, i.e., y = - 2. If x = 0 and y = - 2, then both equations (1) and (2) are satisfied. Therefore the point (0, - 2) is a critical point.

Case 2: Let y = - 3x/2. Then we know that equation (2) is satisfied. We plug y = - 3x/2 into equation (2) and simplify:

x² - 2(- 3x/2) - 4 = 0

x² + 3x - 4 = 0

(x - 1)(x + 4) = 0

x = 1 or x = - 4.

So, we have two solutions of equation (2) for case 2. The first solution is when x = 1, which means y = - 3x/2 = - 3/2. If x = 1 and y = - 3/2, then both equations (1) and (2) are satisfied. Therefore the point (1, - 3/2) is a critical point.

The second solution for case 2 is when x = - 4, which means y = - 3x/2 = 6. Therefore, the point (- 4, 6) is a critical point.

To summarize the results from both case 1 and case 2, we conclude that f (x, y) has three critical points: (0, - 2), (1, - 3/2), and (- 4, 6).

You should double check that Df (x, y) = [0    0] at each of these points.

Step 2: Classify the critical points.

The Hessian matrix is

$$\left[\vphantom{ \begin{array}{cc} 6x+2y & 2x 2x & -2 \end{array} }\right. \begin{array}{cc} 6x+2y & 2x 2x & -2 \end{array} \left.\vphantom{ \begin{array}{cc} 6x+2y & 2x 2x & -2 \end{array} }\right]$$

We need to check the definiteness of the Hf (x, y) at the critical points (0, - 2), (1, - 3/2), and (- 4, 6).

For the critical point (0, - 2),

$$\left[\vphantom{ \begin{array}{rr} -4 & 0 0 & -2 \end{array} }\right. \begin{array}{rr} -4 & 0 0 & -2 \end{array} \left.\vphantom{ \begin{array}{rr} -4 & 0 0 & -2 \end{array} }\right]$$

h₁₁ = - 4 < 0 and det(Hf )= 8 > 0. This means Hf (0, - 2) is negative definite and f has a local maximum at (0, - 2).

For the critical point (1, - 3/2),

$$\left[\vphantom{ \begin{array}{rr} 3 & 2 2 & -2 \end{array} }\right. \begin{array}{rr} 3 & 2 2 & -2 \end{array} \left.\vphantom{ \begin{array}{rr} 3 & 2 2 & -2 \end{array} }\right]$$

h₁₁ = 3 > 0 and det(Hf )= - 6 - 4 = - 10 < 0. This means Hf (1, - 3/2) is indefinite and f has a saddle at (1, - 3/2).

For the critical point (- 4, 6),

$$\left[\vphantom{ \begin{array}{rr} -12 & -8 -8 & -2 \end{array} }\right. \begin{array}{rr} -12 & -8 -8 & -2 \end{array} \left.\vphantom{ \begin{array}{rr} -12 & -8 -8 & -2 \end{array} }\right]$$

h₁₁ = - 12 < 0 and det(Hf )= 24 - 64 = - 40 < 0. This means Hf (- 4, 6) is indefinite and f has a saddle at (- 4, 6).

Example 2

Identify the local extrama of f (x, y) = (x² + y²)e^-y.

Solution

Step 1: Find the critical points.

The derivative of f is

$$\left[\vphantom{2xe^{-y} \quad (2y-x^2-y^2)e^{-y}}\right. \left.\vphantom{2xe^{-y} \quad (2y-x^2-y^2)e^{-y}}\right]$$

Df (x, y) = [0    0] means that 2x = 0 and 2y - x² - y² = 0, i.e., x = 0 and y(2 - y) = 0.

The critical points are therefore (0, 0) and (0, 2).

Step 2: Classify the critical points.

The Hessian matrix is

$$\left[\vphantom{ \begin{array}{cc} 2e^{-y} & -2x e^{-y} -2x e^{-y} & (2-4y +y^2 +x^2) e^{-y} \end{array} }\right. \begin{array}{cc} 2e^{-y} & -2x e^{-y} -2x e^{-y} & (2-4y +y^2 +x^2) e^{-y} \end{array} \left.\vphantom{ \begin{array}{cc} 2e^{-y} & -2x e^{-y} -2x e^{-y} & (2-4y +y^2 +x^2) e^{-y} \end{array} }\right]$$

At the critical point (0, 0)

$$\left[\vphantom{ \begin{array}{rr} 2 & 0 0 & 2 \end{array} }\right. \begin{array}{rr} 2 & 0 0 & 2 \end{array} \left.\vphantom{ \begin{array}{rr} 2 & 0 0 & 2 \end{array} }\right]$$

h₁₁ = 2 > 0 and det(Hf )= 4 > 0, so (0, 0) is a local mininum.

At the critical point (0, 2)

$$\left[\vphantom{ \begin{array}{rr} e^{-2} & 0 0 & -2 e^{-2} \end{array} }\right. \begin{array}{rr} e^{-2} & 0 0 & -2 e^{-2} \end{array} \left.\vphantom{ \begin{array}{rr} e^{-2} & 0 0 & -2 e^{-2} \end{array} }\right]$$

h₁₁ = e^-2 > 0 and det(Hf )= - 2e^-4 < 0 so (0, 2) is a saddle point.