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Partial differentiation and stationary points Watch

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    Hi all.

    I've been asked to find the stationary point of the function u(x,y) = 2x^4 + 8(x^2)y^2 - 4x^2 + 4y^2 .

    I know the stationary points lie at the solution of du/dx = du/dy = 0.

    So I have found
    du/dx = 8x^3 + 16x(y^2) - 8x, and
    du/dy = 16(x^2)y + 8y.

    Presumable I have to set these 2 equations to =0 and solve simultaneously, but am not sure exactly how to do this. Can anyone advise?

    Thanks!
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    That's right. Start by looking at your equation for du/dy. Factorise it and set it equal to zero. You should find that this only works for one value of y. Now take this y value and substitute it into your equation for du/dx. Factorise this and make it equal to zero. You should find three possible x values.
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    Good so far

    du/dx=8x^3 + 16x.y^2 - 8x - eqn1
    du/dy=16y.x^2+8y - eqn2

    check

    du/dxdy=32xy
    du/dydx=32xy


    You put du/dx=0 and du/dy=0 then simplify and you can get values for x and y then put an x value from du/dx into du/dy to find a y-value and vice versa.

    eqn1 - 8x^3 + 16x.y^2 - 8x= 0 => 8x(x^2+2y^2-1)=0
    eqn2 - 16y.x^2+16y = 0 => 16y(x^2+1)=0
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    Cheers guys, will rep tomorrow. I worked through the rest of the problem and obtained 3 stationary points; a saddle at (0,0) and 2 minima at (1,0) and (-1,0), which I think is the correct answer.
 
 
 
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