How Do You Determine the Extrema of f(x, y) = x + 2y on a Unit Disk?

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To determine the extrema of the function f(x, y) = x + 2y on the unit disk defined by x² + y² ≤ 1, the parameterization x = cos(t) and y = sin(t) is used, leading to g(t) = cos(t) + 2sin(t). The derivative g'(t) is set to zero, resulting in the equation 2cos(t) - sin(t) = 0, which simplifies to tan(t) = 2. The next step involves finding the values of t that satisfy this equation, which can be approached by expressing sin(t) and cos(t) in terms of tan(t). The discussion highlights the need for further steps to find the maximum and minimum values of f on the disk.
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Find the maximum and minimum values of f(x,y) = x+2y on the disk
x2+y^2 ≤1

I have this for now:

f_1(x,y) = 1
f_2(x,y) = 2

x=cos(t) and y=sin(t)

I have that g(t) = x(t) + 2*y(t) --> g(t) = cost(t) + 2*sin(t)

g'(t) = 0 = 2*cost-sin(t)

Then I can see that:

2cos(t)/cos(t) -sin(t)/cos(t) = 0/cos(t) --> tan = 2

That is the parameterization, right?

From this point I have no idea what to do.
 
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Kork said:
Find the maximum and minimum values of f(x,y) = x+2y on the disk
x2+y^2 ≤1

I have this for now:

f_1(x,y) = 1
f_2(x,y) = 2

x=cos(t) and y=sin(t)

I have that g(t) = x(t) + 2*y(t) --> g(t) = cost(t) + 2*sin(t)

g'(t) = 0 = 2*cost-sin(t)

Then I can see that:

2cos(t)/cos(t) -sin(t)/cos(t) = 0/cos(t) --> tan = 2

That is the parameterization, right?

From this point I have no idea what to do.

Find the value or values of t that give tan(t) = 2. Or, since you only need sin(t) and cos(t), why not express them in terms of tan(t)?

RGV
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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