Functiosn with multiple variables

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Homework Help Overview

The discussion revolves around functions of multiple variables, specifically focusing on the properties and behaviors of a C1-function and optimization of a given function over specified domains.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the application of the chain rule for differentiating composite functions and question the meaning of specified domains for optimization problems.
  • There is an attempt to clarify the definitions of the domains D1 and D2, as well as the implications for finding extrema.
  • One participant raises a question about the relationship between the gradient and level curves, seeking to understand critical points and the absence of a maximum value.

Discussion Status

Some participants have made progress in applying differentiation techniques and identifying critical points. However, confusion remains regarding the interpretation of the domains for optimization and the relationship between gradients and level curves. Guidance has been offered to clarify the definitions of the domains.

Contextual Notes

Participants are working within the constraints of homework assignments, which may limit the information available for problem-solving. There is an ongoing exploration of assumptions related to the behavior of functions within specified intervals.

Niles
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Homework Statement



1) I have a C1-function f(u,v), and f(0,0) = 1, df/du(0,0) = 3 and df/dv(0,0)=5. I have to find k'(1), when k(x) = f(x^2-1;x-1).

2) A function g(x,y) = 3x^2-x-y+y^2. I have to find the minimum and maximum of the function on D_1 = [0,1] x [0,1] and D_2 = [1,2] x [0,1].

The Attempt at a Solution



1) g(x,y) = x^2-1 and h(x,y) = x-1. dk/dx = (df/du)*(dg/dx) + (df/dv)*(dh/dx). I know g and h, but I have to find f - how do I do that?

2) I know how to find maximum and minimum of a function, but I don't know what they mean by "on D_1 = [0,1] x [0,1] and D_2 = [1,2] x [0,1]."?

EDIT: Sorry for the spelling-error in the title.
 
Last edited:
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I solved #1 - simply by using the chain rule:

dk/dx = (df/du)(g(x),h(x))*(dg/dx) + (df/dv)(g(x),h(x))*(dh/dx)

In my book, the arguments weren't included.. that confused me.

For #2, I have found the critical points as usual, and the minimum is (1/6;1/2). But still, the D_1 and D_2 confuses me - what do they mean by that?
 
Last edited:
Niles said:
2) I know how to find maximum and minimum of a function, but I don't know what they mean by "on D_1 = [0,1] x [0,1] and D_2 = [1,2] x [0,1]."?


These refer to two domains. D1 is the set [itex]\{(x,y):x\in[0,1]\ \text{and}\ y\in[0,1]\}[/itex]. D2 is the set [itex]\{(x,y):x\in[1,2]\ \text{and}\ y\in[0,1]\}[/itex].
 
Ok, thanks. I have another question, but this is about the gradient of a function.

Take a look at the attached level curve of a function. At which point is the gradient 0?
I believe it is C, since the gradient there is 0 - but apparently it is A - why is that?

And btw, to find the critical points in the two sets, I can look at the function and the interval, and see what f(x,y) = 3x^2-x-y+y^2 has to be less than or equal and what is has to be bigger than or equal, and solve it from there?

I get that the minimum is -1/3 - I get that there is no maximum, which I don't understand. I just find the gradient, and equal it to zero - I get one result, and by processing it, I get the minimum-point. How come there is no maximum?

Thank you for all your help so far.
 

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Last edited:
Or perhaps I should "extend" the question and ask, what the link between the gradient and the level curve is.

+ the other question about the minimum/maximum.
 

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