Question

need help on lagrange multiplier problem.

Q: Find the maximum and minimum values of the function f(x; y) = x^2y on the ellipse 4x2+9y2 =36. Explain the geometry of the solution by plotting the ellipse and looking at contours of the function f.

Answer 1

opps the equation of the ellipse 36=4x^2+9y^2

Answer 2

Rbrass, you have to take the derivative implicitly, since the function is defined in term of both variables. As for the geometry and contours, look at your resulting equation to determine that.

Answer 3

i know the gradient of f(x,y)= the gradient of g(x,y)* a constant so from that i got
<2xy,x^2>=k<8x,18y> where k is some constant

Answer 4

okay any idea on how to solve for the min/max points? i got the equations
2xy=k8x
x^2=k18y
0=4x^2+9y^2+36

Answer 5

i mean 0=4x^2-9y^2-36

Answer 6

don't really know where to go from there

Answer 7

Rbrass33, is that second function \[f(x,y) = x ^{2y}\]? for real? lol

Answer 8

oh no its (x^2)*y sorry about that

Answer 9

\[f(x,y) = yx^2\]

Answer 10

yup thats what i meant to put

Answer 11

This one's beyond me; sorry I can't help for now

Answer 12

haha its okay thanks for trying

Answer 13

First, it helps to get a visual before we dive into the gnarly algebra:
go to
http://www.wolframalpha.com
and type
extrema of x^2y on 36=4x^2+9y^2

Answer 14

oh nice i didn't know you could do that on wolfram

Answer 15

Here are some great notes on lagrange multipliers:
http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx

Answer 16

okay thanks i found 3 equations but i still don't have enough information to solve them
2xy=k8x
x^2=k18y
0=4x^2+9y^2-36

Answer 17

ahhh figured it out...thanks mathematica