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calling the calculus 3 gods

O

Oscar1613

Golden Member
my teacher gave us a b!tch of a problem on our last test and since no one got it 100% right, he told us to try to figure it out for homework... the more i try, the uglier this problem looks... PLEASE tell me theres an easy way to do this problem that i'm not seeing...

Find the absolute max/min values of f(x,y)=e^(-x^2 - y^2)*(x^2 + 2y^2) on the disk D with x^2 + y^2 <= 4

😕 any help is very much appreciated 🙂
 
You may have better luck in the Highly Technical forum.

The math gods of Anandtech tend to frequent there moreso than Off-Topic. 🙂
 
Originally posted by: N8Magic
You may have better luck in the Highly Technical forum.

The math gods of Anandtech tend to frequent there moreso than Off-Topic. 🙂
Click to expand...

thanks, i'll try there...
 
The max/min points are when both Fx and Fy are zero. Then you do the d test to find out which is which:
|Fxx Fyx| = d
|Fxy Fyy|

Test d at the point(s) you arrive at. I forgot which values of d's indicate max's, min's, saddle pts etc.

Edit, make sure the points you are testing are within 2 units from the origin because of the constraint.
 
f(x,y) converted to f(r) yields f(r)=e^(-r^2)*(r^2). Now I'm not sure if you meant that to be e raised to the whole thing or if it was e raised to negative r squared. You need to clarify.
 
Are you sure that's (rewritten) f(x,y) = (x^2 + 2y^2) * e^(-x^2 - y^2) and not f(x,y) = (x^2 + y^2) * e^(-x^2 - y^2)
 
i know the fx fy=0 method, but that gets very messy very fast, so i thought i might be missing something...

i also know how to convert to polar coordinates, but we never had to use them in a find the max/min problem, so i'm not sure how i would do everything with it...

lagrange wouldn't work here, right?
 
f'(r) = -2r*e^(-r^2) + 2r*e^(-r^2)

Wait, doesn't that add to zero? Does that mean that any point is a rel max/min? Maybe it's a planar object. Try graphing off the ti-89.


Edit, scratch that above cuz there's a 2 by the y^2, so you can't directly go from x^2 and y^2 to r^2.
 
Speaking strictly as someone who is Mathematically Challenged; what the fvck is the point to an equation like this?

Can you build a bridge with it? Can you cure cancer? Can you find out EXACTLY how many apples Suzy has in her cart, being that Matthew's train left NYC four hours ago, traveling at an average rate of 34.13 KM/s?

I mean, really! That's just a SICKASS equation!
 
what the fvck is the point to an equation like this?
Click to expand...

Hehe, well I get to rack up major 6.0 HPA credits 🙂

Oh ya, lagringe doesn't work here cuz that's for putting two constraints together (ie finding maximum box volume inside an ellipsoid)
 
Originally posted by: Marty
Are you sure that's (rewritten) f(x,y) = (x^2 + 2y^2) * e^(-x^2 - y^2) and not f(x,y) = (x^2 + y^2) * e^(-x^2 - y^2)
Click to expand...

yes the first one is correct... the (x^2 + 2y^2) is outside of the e^
 
Originally posted by: Ionizer86
f'(r) = -2r*e^(-r^2) + 2r*e^(-r^2)

Wait, doesn't that add to zero? Does that mean that any point is a rel max/min? Maybe it's a planar object. Try graphing off the ti-89.
Click to expand...

you forgot about theta...
 
Originally posted by: MichaelD
Speaking strictly as someone who is Mathematically Challenged; what the fvck is the point to an equation like this?

Can you build a bridge with it? Can you cure cancer? Can you find out EXACTLY how many apples Suzy has in her cart, being that Matthew's train left NYC four hours ago, traveling at an average rate of 34.13 KM/s?

I mean, really! That's just a SICKASS equation!
Click to expand...

no kidding... now imagine seeing that problem on a test when it was worth 1/3rd of the test...
 
Well, the 2 in front of the y^2 is what makes it difficult; there is no point in going to polar coordinates because of it. I am not sure there is much you can do besides just churn through it.
 
Originally posted by: Marty
Well, the 2 in front of the y^2 is what makes it difficult; there is no point in going to polar coordinates because of it. I am not sure there is much you can do besides just churn through it.
Click to expand...

uggggh.... multiplying those double derivatives is going to be really messy....
 
All right, the TI-89 makes stuff easy. Fx = -2x(x^2 + 2y^2 +1) * e^(-x^2 - y^2)
Fy = -2y(2y^2 + x^2 -2) * e^(-y^2 - x^2) Now to set for zero.... (how do you do this on a TI-89 again?)
 
You have to use lagrange multipliers.

f(x,y)x = (lambda)g(x,y)x
f(x,y)y = (lambda)g(x,y)y

Then find the x values and y values that fit the equation. Plug them into g(x,y) and solve them. Then once you have your x and y values, plug them into the f(x,y) function to see which are min and which are max. Max have the highest values and min have the lowest values.

g(x,y) = x^2 + y^2 <= 4
f(x,y)x is the partial derivative of f(x,y) with respect to x.
g(x,y)x is the partial derivative of the function g(x,y) with respect to x.

same with f(x,y)y and g(x,y)y.


Edit: Oh yeah finding the partial derivatives is going to be a b!tch 😀
 
Originally posted by: DAWeinG
You have to use lagrange multipliers.

f(x,y)x = (lambda)g(x,y)x
f(x,y)y = (lambda)g(x,y)y

Then find the x values and y values that fit the equation. Plug them into g(x,y) and solve them. Then once you have your x and y values, plug them into the f(x,y) function to see which are min and which are max. Max have the highest values and min have the lowest values.

g(x,y) = x^2 + y^2 <= 4
f(x,y)x is the partial derivative of f(x,y) with respect to x.
g(x,y)x is the partial derivative of the function g(x,y) with respect to x.

same with f(x,y)y and g(x,y)y.
Click to expand...
i thought lagrange might be part of it, but someone in my class said he hinted not to use that... and he usually puts that in the directions for the problem if we need to use it...

i feel like this guy right about now....
 
just tried lagrange... doesnt seem to work, unless i did something wrong...

2x*e^(-x^2 - y^2)*(1 - x^2 - 2y^2) = (lamda)*2x
2y*e^(-x^2 - y^2)*(2 - x^2 - 2y^2) = (lamda)*2y

if x != 0 and y !=0 then
e^(-x^2 - y^2)*(1 - x^2 - 2y^2) = (lamda)
e^(-x^2 - y^2)*(2 - x^2 - 2y^2) = (lamda)
so then
(1 - x^2 - 2y^2)=(2 - x^2 - 2y^2)
which doesnt work...😕


*dies*
 
Ok, so Gx = 2x and Gy = 2y

Now 2x ^ = -2x(x^2 + 2y^2 + 1) * e^(-x^2 - y^2)
^ = -(x^2 + 2y^2 + 1) * e^(-x^2 - y^2)

2y ^ = -2y(2y^2 + x^2 - 2) * e^(-y^2-x^2)
^ = -(2y^2 + x^2 - 2) * e^(-y^2-x^2)

(2y^2 + x^2 - 2)*e^(-y^2-x^2) = (x^2+2y^2+1) * e^(-x^2 - y^2)
2y^2 + x^2 - 2 = x^2 + 2y^2 +1
0 = 3

(^ = lambda) Let's hope the math is correct.
Doesn't work either though you may want to check the work

BTW, the TI-89 points out 0,0,0 as a min, and there's no max points cuz it's like a paraboloid.
 
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