Section 3.7 Maxima and Minima

3.7 Maxima and Minima

In calculus I, the most useful application for taking derivative of a function is finding the maximum and the minimum of the function. In this section, we learn how to find maxima or minima of a two variables function. First, we define what does it mean to have a maximum or a minimum at a point for a two variable function.

Definition: 

Let $z=f(x,y)$ be a function of two variables that is defined and continuous on an open set containing the point $(x_0,y_0)$. Then $f$ has a local maximum (local minimum) at $(x_0,y_0)$ if 

$$f(x_0,y_0)\ge f(x,y),(f(x_0,y_0)\le f(x,y))$$

for all points $(x,y)$ within some disk centered at $(x_0,y_0)$. The number $f(x_0,y_0)$ is called a local maximum value (local minimum value). If the preceding inequality holds for every point $(x,y)$ in the domain of $f$ , then $f$ has a global maximum/absolute maximum (global minimum/absolute minimum) at $(x_0,y_0)$. 

Definition: critical point 

Let $z=f(x,y)$ be a function of two variables that is defined on an open set containing the point $(x_0,y_0)$. The point $(x_0,y_0)$ is called a critical point of a function of two variables $f$ if one of the two following conditions holds: 

1. $f_x(x_0,y_0)=f_y(x_0,y_0)=0$ 

2. Either $f_x(x_0,y_0)$ or $f_y(x_0,y_0)$ does not exist.

Example 1: Find the critical points of the functions: $z=f(x,y)=x^2+xy+y^2+y+4$. 

Exercise 1: Find the critical points of the functions: $z=f(x,y)=x^2-3xy+y^2+3y+1$. 

Theorem: Fermat\textquoteright s Theorem for Functions of Two Variables

Let $z=f(x,y)$ be a function of two variables that is defined and continuous on an open set containing the point $(x_0,y_0)$. Suppose $f_x$ and $f_y$ each exists at $(x_0,y_0)$. If $f$ has a local extremum at $(x_0,y_0)$, then $(x_0,y_0)$ is a critical point of $f$. 

Definition: Given the function $z=f(x,y)$, the point $(x_0,y_0,f(x_0,y_0))$ is a saddle point if both $f_x(x_0,y_0)=f_y(x_0,y_0)=0$, but $f$ does not have a local extremum at $(x_0,y_0)$.

Theorem: Second Derivative Test 

Let $z=f(x,y)$ be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point $(x_0,y_0)$. Suppose $f_x(x_0,y_0)=f_y(x_0,y_0)=0$.

Define the quantity

$$D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0){)}^2.$$

i. If $D>0$ and $f_{xx}(x_0,y_0)>0$, then $f$ has a local minimum at $(x_0,y_0)$. 

ii. If $D>0$ and $f_{xx}(x_0,y_0)<0$, then $f$ has a local maximum at $(x_0,y_0)$. 

iii. If $D<0$, , then $f$ has a saddle point at $(x_0,y_0)$.

iv. If $D=0$, then the test is inconclusive.

Example 2: Find the critical points of the function $f(x,y)=x^2+3y^2+2x-12y+4$, and use the second derivative test to find the local extrema.

Exercise 2: Find the critical points of the function $f(x,y)=2x^2-y^2+12x+8y-3$, and use the second derivative test to find the local extrema.

Example 3: Find the critical points of the function $f(x,y)=\frac{1}{3}{x}^3+y^2+2xy-6x-3y+4$, and use the second derivative test to find the local extrema.

Exercise 3: Find the critical points of the function $f(x,y)=x^3+4xy-8x-4y^2$, and use the second derivative test to find the local extrema.

Theorem: Extreme Value Theorem 

A continuous function $f(x,y)$ on a closed and bounded set $D$ in the plane attains an absolute maximum value at some point of $D$ and an absolute minimum value at some point of $D$.

Theorem: Finding Extreme Values of a Function of Two Variables 

Assume $z=f(x,y)$ is a differentiable function of two variables defined on a closed, bounded set $D$. Then $f$ will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following: 

i. The values of $f$ at the critical points of $f$ in $D$. 

ii. The values of $f$ on the boundary of $D$.

Example 4: Find absolute extrema of the function $f(x,y)=x^2+2y^2-4x-4y+2$ on the domain defined by $0\le x\le 4$ and $-2\le y\le 2$.

Exercise 4: Find absolute extrema of the function $f(x,y)=4x^2+y^2-8x+2y+3$ on the domain defined by $0\le x\le 2$ and $-1\le y\le 3$.

Example 5: Find absolute extrema of the function $f(x,y)=x^2+y^2-2x+4y$ on the range $R=\{(x,y)|x^2+y^2\le 9\}.$

Exercise 5: Find absolute extrema of the function $f(x,y)=x^2+y^2-2y+1$ on the range $R=\{(x,y)|x^2+y^2\le 4\}.$

Group work:

1. Find absolute extrema of the function $f(x,y)=10-6x+x^2-4y+y^2$ on the range $R$ is the triangular region with vertices $(0,0),(4,0),$ and $(4,4)$.

2. Find absolute extrema of the function $f(x,y)=x^2{y}^2$ on the range $R=\{(x,y)|x\ge 0,x^2+y^2\le 4\}.$

3. Find the critical points of the function $f(x,y)=x^3+y^3-300x-75y-3$, and use the second derivative test to find the local extrema.

4. Find absolute extrema of the function $f(x,y)=3x^2-2xy+y^2-8y$ on the domain defined by $-2\le x\le 2$ and $0\le y\le 4$.

5. Find absolute extrema of the function $f(x,y)=(x-2{)}^2+(y-4{)}^2$ on the range $R$ is the triangular region with vertices $(0,0),(4,0),$ and $(0,4)$.