# Analyzing Critical Points of f(x,y)=Ax2+E

• xokaitt
In summary, by finding the first and second partial derivatives of f(x,y), we can determine that all points lying on the y-axis are critical points with a value of 0 for the second derivative. By graphing the surface, we can see that these critical points are saddle points. Therefore, we can conclude that the critical points of f(x,y) are saddle points.
xokaitt

## Homework Statement

Let f(x,y)=Ax2+E where A and E are constants. What are the critical points of f(x,y)? Determine whether the critical points are local maxima, local minema, or saddle points.

2. The attempt at a solution

First I found the first partial derivatives with respect to x and y
$$\partial$$f/$$\partial$$x=2Ax
$$\partial$$f/$$\partial$$y=0
$$\Rightarrow$$ 2Ax=0,
$$\Rightarrow$$ x=0 for any constant A.

Therefore, all points lying on the y-axis are critical points.
(i.e. C.P.'s = (0,n), n$$\in$$R.)

Now, we have to find the second partial's with respect to x and y.
$$\partial$$2f/$$\partial$$x2=2A
$$\partial$$2f/$$\partial$$y2=0
and
$$\partial$$2f/$$\partial$$x$$\partial$$y=0

Therefore Df=($$\partial$$2f/$$\partial$$x2)($$\partial$$2f/$$\partial$$y2)-($$\partial$$2f/$$\partial$$x$$\partial$$y)2 at (0,n) , n$$\in$$R.
$$\Rightarrow$$ Df=(2A)(0)-(0)2=0

This is where I get stuck. Now that Df=0, how do I determine whether or not the critical pts are local extrema or saddle pts?

From plotting the function on Mathematica, I know that these critical points are in fact saddle points, but I don't know how to mathematically state that.

Thanks!

It seems to me that you are making this much harder than it needs to be by not sketching a graph of this surface. Since y doesn't appear explicitly in the formula for the function, this surface is a cylinder with parabolic cross section, and with its axis of symmetry in the direction of the y-axis. IOW, the surface looks something like a trough. If A > 0, the trough opens upward, and all critical points are global minima. If A < 0, the trough opens downward, and all critical points are global maxima. All critical points lie on a line that is parallel to the y-axis.

## 1. What is the purpose of analyzing critical points?

The purpose of analyzing critical points is to determine the behavior and properties of a function, such as its maximum and minimum values, inflection points, and saddle points. This information is useful in understanding the overall behavior of the function and can aid in solving optimization problems.

## 2. How do you find the critical points of a function?

To find the critical points of a function, you first take the partial derivatives of the function with respect to each variable. Then, set these partial derivatives equal to 0 and solve for the values of x and y that make them true. These values represent the critical points of the function.

## 3. What is the significance of the A and E values in the function f(x,y)=Ax2+E?

The A value represents the coefficient of the x2 term in the function, which determines the shape of the function. A positive A value results in a parabola opening upwards, while a negative A value results in a parabola opening downwards. The E value represents the constant term in the function, which shifts the entire graph up or down along the y-axis.

## 4. Can a function have more than one critical point?

Yes, a function can have multiple critical points. This occurs when there are multiple values of x and y that satisfy the equations obtained from taking the partial derivatives and setting them equal to 0. These critical points can have different properties, such as being a maximum or minimum point, or a saddle point.

## 5. How can critical points be used to solve optimization problems?

Critical points can be used to solve optimization problems by identifying the maximum or minimum values of a function. These values correspond to the critical points of the function, and can be found by analyzing the first and second partial derivatives of the function at these points. By finding the critical points and evaluating the function at these points, the optimal solution to the optimization problem can be determined.

• Calculus and Beyond Homework Help
Replies
8
Views
316
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
22
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
921
• Calculus and Beyond Homework Help
Replies
1
Views
755
• General Math
Replies
1
Views
891
• Calculus and Beyond Homework Help
Replies
9
Views
3K
• Calculus and Beyond Homework Help
Replies
9
Views
3K
• Calculus and Beyond Homework Help
Replies
5
Views
6K