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OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

don't really know where to go from there

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yup thats what i meant to put

OpenStudy (anonymous):

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

OpenStudy (anonymous):

haha its okay thanks for trying

OpenStudy (mathteacher1729):

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

OpenStudy (anonymous):

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

OpenStudy (mathteacher1729):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ahhh figured it out...thanks mathematica

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