Critical points, several variables

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The critical points of the function x^3 + y^3 + 3x^2 + 6y^2 - 9x + 9y + 1 are determined by setting the partial derivatives df/dx and df/dy to zero. The solutions for df/dx yield x values of -3 and 1, while df/dy gives y values of -3 and 1. This results in four critical points: (-3, -3), (-3, 1), (1, -3), and (1, 1). The discussion highlights the challenge of finding corresponding y values for given x values and vice versa. Overall, the critical points are identified without needing to define them further.
simba_
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Homework Statement


Find the critical points of

x3 + y3 + 3x2 + 6y2 - 9x + 9y +1

you do not need to define the critical points

Homework Equations





The Attempt at a Solution



i have
df/dx = 3x2 + 6x - 9 and when i solve this x = -3, 1
but i don't know what the corresponding y values are

df dy = 3y2 + 12y + 9 and so y = -3, -1
and here i don't know what the corresponding x values are
 
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simba_ said:

Homework Statement


Find the critical points of

x3 + y3 + 3x2 + 6y2 - 9x + 9y +1

you do not need to define the critical points

Homework Equations





The Attempt at a Solution



i have
df/dx = 3x2 + 6x - 9 and when i solve this x = -3, 1
but i don't know what the corresponding y values are

df dy = 3y2 + 12y + 9 and so y = -3, -1
and here i don't know what the corresponding x values are
fx = 0 when x = -3 or x = 1
fy = 0 when y = -3 or y = 1

So both partials are zero at (-3, -3), (-3, 1), (1, -3), and (1, 1).
 
tyty
 
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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