Multivariable calculus and vector analysis

Extrema examples

Example 1

Find the local extrema of $f\left(x,y\right)=x^3+x^{2}y-y^2-4y$.

Solution

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

$$\begin{array}{lll}\hfill Df\left(x,y\right)=\left[3x^2+2xy{x}^2-2y-4\right].& & \hfill \end{array}$$

$Df\left(x,y\right)=\left[00\right]$ means both components must be zero simultaneously. We need

$$\begin{array}{lll}\hfill x\left(3x+2y\right)=0& & \hfill \text{ (1)}\end{array}$$

and

$$\begin{array}{lll}\hfill x^2-2y-4=0.& & \hfill \text{ (2)}\end{array}$$

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 $\left(0,-2\right)$ 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:

$$\begin{array}{llll}\hfill x^2-2\left(-3x∕2\right)-4& =0& \hfill & \\ \hfill x^2+3x-4& =0& \hfill & \\ \hfill \left(x-1\right)\left(x+4\right)& =0& \hfill & \\ \hfill x=1\text{ or }x& =-4.& \hfill & \end{array}$$

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 $\left(1,-3∕2\right)$ is a critical point.

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

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

You should double check that $Df\left(x,y\right)=\left[00\right]$ at each of these points.

Step 2: Classify the critical points.

The Hessian matrix is

$$\begin{array}{lll}\hfill Hf\left(x,y\right)=\left[\begin{array}{cc}\hfill 6x+2y\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill -2\hfill \end{array}\right]& & \hfill \end{array}$$

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

For the critical point $\left(0,-2\right)$,

$$\begin{array}{lll}\hfill Hf\left(0,-2\right)=\left[\begin{array}{cc}\hfill -4& \hfill 0\\ \hfill 0& \hfill -2\end{array}\right]& & \hfill \end{array}$$

$h_{11}=-4<0$ and $det\left(Hf\right)=8>0$. This means $Hf\left(0,-2\right)$ is negative definite and $f$ has a local maximum at $\left(0,-2\right)$.

For the critical point $\left(1,-3∕2\right)$,

$$\begin{array}{lll}\hfill Hf\left(1,-3∕2\right)=\left[\begin{array}{cc}\hfill 3& \hfill 2\\ \hfill 2& \hfill -2\end{array}\right].& & \hfill \end{array}$$

$h_{11}=3>0$ and $det\left(Hf\right)=-6-4=-10<0$. This means $Hf\left(1,-3∕2\right)$ is indefinite and $f$ has a saddle at $\left(1,-3∕2\right)$.

For the critical point $\left(-4,6\right)$,

$$\begin{array}{lll}\hfill Hf\left(-4,6\right)=\left[\begin{array}{cc}\hfill -12& \hfill -8\\ \hfill -8& \hfill -2\end{array}\right].& & \hfill \end{array}$$

$h_{11}=-12<0$ and $det\left(Hf\right)=24-64=-40<0$. This means $Hf\left(-4,6\right)$ is indefinite and $f$ has a saddle at $\left(-4,6\right)$.

Example 2

Identify the local extrama of $f\left(x,y\right)=\left(x^2+y^2\right)e^{-y}$.

Solution

Step 1: Find the critical points.

The derivative of $f$ is

$$\begin{array}{lll}\hfill Df\left(x,y\right)=\left[2x{e}^{-y}\left(2y-x^2-y^2\right)e^{-y}\right]& & \hfill \end{array}$$

$Df\left(x,y\right)=\left[00\right]$ means that $2x=0$ and $2y-x^2-y^2=0$, i.e., $x=0$ and $y\left(2-y\right)=0$.

The critical points are therefore $\left(0,0\right)$ and $\left(0,2\right)$.

Step 2: Classify the critical points.

The Hessian matrix is

$$\begin{array}{lll}\hfill Hf\left(x,y\right)=\left[\begin{array}{cc}\hfill 2e^{-y}\hfill & \hfill -2x{e}^{-y}\hfill \\ \hfill -2x{e}^{-y}\hfill & \hfill \left(2-4y+y^2+x^2\right)e^{-y}\hfill \end{array}\right]& & \hfill \end{array}$$

At the critical point $\left(0,0\right)$

$$\begin{array}{lll}\hfill Hf\left(0,0\right)=\left[\begin{array}{cc}\hfill 2& \hfill 0\\ \hfill 0& \hfill 2\end{array}\right]& & \hfill \end{array}$$

$h_{11}=2>0$ and $det\left(Hf\right)=4>0$, so $\left(0,0\right)$ is a local mininum.

At the critical point $\left(0,2\right)$

$$\begin{array}{lll}\hfill Hf\left(0,2\right)=\left[\begin{array}{cc}\hfill e^{-2}& \hfill 0\\ \hfill 0& \hfill -2e^{-2}\end{array}\right]& & \hfill \end{array}$$

$h_{11}=e^{-2}>0$ and $det\left(Hf\right)=-2e^{-4}<0$ so $\left(0,2\right)$ is a saddle point.

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

2a. The partial derivative

2b. The derivative and differentiability

2c. The chain rule, directional derivative and gradient

2d. Applications of differential calculus

• Taylor's theorem
• Taylor polynomial example
• Local extrema
• Extrema examples

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material