Finding the maximum and minimum values

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SUMMARY

The discussion focuses on finding the maximum and minimum values of the function f(x,y)=x+2y constrained by the disk defined by x^2+y^2 ≤ 1. The initial steps involve calculating the partial derivatives fx(x,y) = 1 and fy(x,y) = 2, indicating that there are no critical points within the disk. The solution requires evaluating the function on the boundary of the disk, specifically substituting y=±√(1-x^2) into f(x,y) and differentiating to find where the derivative equals zero. Additionally, the values of f at the boundary points (-1, 0) and (1, 0) must be considered to determine the extrema.

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Kork
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I am having trouble with getting started with this one:

Find the maximum and minimum values of f(x,y)=x+2y on the disk x^2+y^2 ≤ 1

I have started like this:
fx(x,y) = 1
fy(x,y) = 2

and then I am lost...How do I solve it?
 
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well, seeing as f(x,y)=0 is simply a line, you can find the intersection points between the line y=-1/2x, and the boundary of the disk, therefore the circle with radius=1.

If you consider f(x,y) as a scalar field, then it is a family of lines with incline=-1/2. Then, you just need to find the constant c (f(x,y)=c), for which the function is tangent to the boundary of the circle.
 
Kork said:
I am having trouble with getting started with this one:

Find the maximum and minimum values of f(x,y)=x+2y on the disk x^2+y^2 ≤ 1

I have started like this:
fx(x,y) = 1
fy(x,y) = 2

and then I am lost...How do I solve it?
Since those are never 0, there is no solution inside the circle. You need to look on the circle. There y= \pm\sqrt{1- x^2} so f(x,y) becomes x+ 2\sqrt{1-x^2} or x- 2\sqrt{1- x^2}. Differentiate those and see where the derivative is 0. Don't forget to look specifically at the value of f at the points (-1, 0) and (1, 0), the endpoints of those intervals.
 

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