# Maximum and Minimum Values (Trig)

1. Oct 17, 2014

### Chas3down

1. The problem statement, all variables and given/known data
Find the 4 critical points f(x,y) = 5ycos(9x) closest to (0,0)

2. Relevant equations

3. The attempt at a solution

fx = -45ysin(9x)
fy = 5cos(9x)
fxx = -45*9ycos(9x)
fyy = 0
fxy = -45sin(9x)

y=0
x=pi/18

(0,pi/18) (0,pi/18 + pi/2) (0,pi/18 - pi/2) (0,pi/18 + 3pi/2) Was not correct.

(All 4 points need to be correct for me to check it )

Last edited: Oct 17, 2014
2. Oct 17, 2014

### Staff: Mentor

Usually the x-coordinate comes first and y second in the notation.
And check the spacing of the solutions in x, it is not pi/2.

3. Oct 17, 2014

### Chas3down

Oops, I tried it correctly first but then swapped to see if that helped with anything, i put it back in x/y

But, i redid the problem and got..
y=0, x = (1/18)pi(4n-1)
y=0, x = (1/18)pi(4n+1)

Not sure how to translate that to the 4 closest points? Tried this but didn't work..

((1/18)pi(4-1),0), ((1/18)pi(0-1),0), ((1/18)pi(4+1),0), ((1/18)pi(0+1),0)

4. Oct 17, 2014

### Staff: Mentor

I don't know how powerful the automatic (?) system to check the points is. I would not try to feed it with calculations like that. Maybe it needs decimal numbers, maybe something like 3/18 pi is okay (but where is the point in things like "4-1"?
((1/18)pi(4+1),0) is not among the 4 closest points.

5. Oct 17, 2014

### Chas3down

Ah, thanks a bunch!
Solution:
((1/18)pi(3),0), ((1/18)pi(-1),0), ((1/18)pi(-3),0), ((1/18)pi(1),0)

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