- #1
Spatulatr0n
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Homework Statement
I was given an assignment to find all relative extrema and saddle points of the equation
f(x,y) = 1/3x^4 + 1/2y^4 - 4xy^2 +2x^2 + 2y^2 + 3
I derived the first partial with respect to x and the first partial with respect to y, but when I tried to find where they both equal 0, the problem became really complicated. I don't know what I am suppose to do. Rather stuck.
Homework Equations
partial x: 4/3x^3 - 4y^2 + 4x
0 = 1/3x^3 - y^2 + x
partial y: 2y^3 - 8xy + 4y
0 = 2y(y^2 - 4x +2)
The Attempt at a Solution
I have tried using substition after setting the equations equal to each other, by using the partial derivatives to find
x = (y^2 + 2)/4
and
y = + or - sqrt(4x - 2)
y = + or - sqrt(1/3x^3 + x)
I have tried setting these two equal to each other to solve for x, but I get a cubic that I don't know what to do with. Is there something like "completing the cube" I could use? Haha.
(1/3x^3 - 3x + 2 = 0)
I'm so confused...please help me. :(