# Max/Min in Calc III- Confused about relation between equations logic?

• giacomh
In summary: XZlLCBZb3VyIGFzc2Vzc21pbmFyeSBvZiB0aGlzIHByb2JsZW0gY3JlYXRlZCAtIHNvIG90aGVyIGZyb20gYSBsb2dpYyBvbmUtYW5zd2VyIGJ1dCBub3QgdGhlIHRvdGFsIHNvbHV0aW9uLiBNZW1vcnkgb2YgYSBsb2dpYyBvbmUtYW5zd2VyIGFuZCBvbmUgb3V0IG9mIHRoZSBmY
giacomh
So, I technically got this problem correct (our professor gave us the answers but not the total solution). My question is more of a logic one- an answer that I'll need to apply this concept to problems on the exam. I included it at the bottom of the post, after I explained the problem

Find local max and min of f(x,y)=xy2-yx2+xy

## Homework Equations

fx=y2-2yx+y fy=2xy-x2+x
fxx=-2y fxy=2y-2x+1 fyy=2x

## The Attempt at a Solution

y2-2yx+y=0
y(y-2x+1)=0
y=0 OR y-2x+1=0

2xy-x2+x
x(2y-x+1)=0
x=0 OR 2y-x+1=0

Plug y=0 into 2y-x+1=0
x=1

Plug x=0 into y-2x+1=0
y=-1

Therefore, -y=x. Plug this into y-2x+1=0

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

Critical Points:
(1,0)
(0,1)
(1/3,-1/3)

**(I know how to solve for min and max from here, I don't need to type it out)

My question:
I don't understand the relationship between the fy and fx derivatives. For instance, I understand why x=0 OR 2y-x+1=0 (from fx), but I don't understand why you would assume that plugging y=0 into 2y-x+1=0 is correct? Why does 2y-x+1=0 when y=0? Rather, why does plugging y=0 into 2y-x+1=0 give you a point on the graph, when 2y-x+1 isn't the derivative itself, only only part of the factored equation? Why don't you take into account the y factored out of (2y-x+1), since you can't factor out a zero (because you're assuming y=0). If x=0 and y=0 are both solutions, why isn't (0,0) a critical point?

Furthermore, I understand the algebra behind plugging -y=x into y-2x+1=0 to get -1/3 and 1/3, but I would never think to do that. How do you know there is a third answer, when you already found 2 answers from the factored y-2x+1=0 (or 2y-x+1=0).

Its just hard for me to visualize what's going on!

giacomh said:
So, I technically got this problem correct (our professor gave us the answers but not the total solution). My question is more of a logic one- an answer that I'll need to apply this concept to problems on the exam. I included it at the bottom of the post, after I explained the problem

Find local max and min of f(x,y)=xy2-yx2+xy

## Homework Equations

fx=y2-2yx+y fy=2xy-x2+x
fxx=-2y fxy=2y-2x+1 fyy=2x

## The Attempt at a Solution

y2-2yx+y=0
y(y-2x+1)=0
y=0 OR y-2x+1=0

2xy-x2+x
x(2y-x+1)=0
x=0 OR 2y-x+1=0

Plug y=0 into 2y-x+1=0
x=1

Plug x=0 into y-2x+1=0
y=-1

Therefore, -y=x. Plug this into y-2x+1=0

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

Critical Points:
(1,0)
(0,1)
(1/3,-1/3)

**(I know how to solve for min and max from here, I don't need to type it out)

My question:
I don't understand the relationship between the fy and fx derivatives. For instance, I understand why x=0 OR 2y-x+1=0 (from fx), but I don't understand why you would assume that plugging y=0 into 2y-x+1=0 is correct? Why does 2y-x+1=0 when y=0? Rather, why does plugging y=0 into 2y-x+1=0 give you a point on the graph, when 2y-x+1 isn't the derivative itself, only only part of the factored equation? Why don't you take into account the y factored out of (2y-x+1), since you can't factor out a zero (because you're assuming y=0). If x=0 and y=0 are both solutions, why isn't (0,0) a critical point?

Furthermore, I understand the algebra behind plugging -y=x into y-2x+1=0 to get -1/3 and 1/3, but I would never think to do that. How do you know there is a third answer, when you already found 2 answers from the factored y-2x+1=0 (or 2y-x+1=0).

Its just hard for me to visualize what's going on!

The conditions f_x = 0 and f_y = 0 give you two equations:
y^2-2yx+y = 0 (1)
2xy-x^2+x = 0 (2)

Eq (1) implies y(y - 2x + 1) = 0, so we must have either
Case (a) y = 0; or Case (b) y - 2x + 1 = 0 --> y = 2x - 1.

Note: at this point we have used eq (1) fully, so now look at eq (2).

In Case (a), eq (2) gives x = x^2, so either x = 0 or x = 1; that is, we have two solutions in case (a): (x,y) = (0,0) and (x,y) = (1,0).

In Case (b) we substitute y = 2x - 1 into eq (2), to get 2x(2x-1)-x^2+x = 0 = x(4x-2-x+1), hence x = 0 or 3x=1. When x = 0 we have y = 2*0-1 = -1 and when x = 1/3 we have y = 2/3 - 1 = -1/3. Therefore, case (b) leads to solutions (x,y) = (0,-1) and (x,y) = (1/3,-1/3).

By breaking the analysis up into cases you can avoid getting mixed up or missing some possibilities.

RGV

## 1. What is the difference between absolute and relative extrema in Calc III?

In Calc III, absolute extrema refer to the highest and lowest points on a graph or function, while relative extrema refer to the local highest and lowest points within a specific interval.

## 2. How do I find the absolute extrema of a multivariable function?

To find the absolute extrema of a multivariable function, you must first take the partial derivatives with respect to each variable and set them equal to 0. Then, solve for the critical points and plug them into the original function to determine if they are absolute extrema.

## 3. Can a function have multiple relative extrema?

Yes, a function can have multiple relative extrema within a given interval. This occurs when the function changes from increasing to decreasing or vice versa.

## 4. What is the meaning of the critical points of a function?

The critical points of a function are the points where the derivative of the function is equal to 0 or does not exist. These points can help us determine the location of extrema and the concavity of the function.

## 5. How do I determine if a critical point is a maximum, minimum, or saddle point?

To determine the type of critical point, we can use the second derivative test. If the second derivative is positive at the critical point, it is a minimum. If the second derivative is negative, it is a maximum. If the second derivative is 0, it is a saddle point.

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