Julius Caro Posted January 19, 2006 Share Posted January 19, 2006 So I have to find the extremum of this function: 2z^3+2z(x+y)+2(x^2+y^2)-1=0 The function is given in an amplicit formula. I tried to plot it with Derive, but failed, hehe. I think there is a maximum (or minimum) in x=y=-1/2, but failed to prove it. If anybody can plot the damn thing OR know how to find maximum/minimum in this function, please guide me!! Link to comment Share on other sites More sharing options...
XgD Posted January 19, 2006 Share Posted January 19, 2006 Go Go Autograph!! Hope this helps XgD Link to comment Share on other sites More sharing options...
Julius Caro Posted January 19, 2006 Author Share Posted January 19, 2006 Thanks a LOT! So it's have a maximum near there, hehe! Link to comment Share on other sites More sharing options...
Julius Caro Posted January 19, 2006 Author Share Posted January 19, 2006 Yeah ! the maximum is there (-0.5,-0.5,1) Thanks :D Link to comment Share on other sites More sharing options...
dreamz Veteran Posted January 19, 2006 Veteran Share Posted January 19, 2006 remember how to do implicit differentiation. your function is z = f(x,y), but defined implicitly as F(x,y,z) = 0. knowing this, you can compute dz/dx and dz/dy. what's the fastest way? differentiation F(x,y,z) = 0 with respect to x using the chain rule: dF/dx*dx/dx + dF/dy*dy/dx + dF/dz*dz/dx = 0 let's rewrite dF/di as Fi for all i. recall that dx/dx = 1 and dy/dx = 0 (because you're holding y constant). this gives you: Fx + Fz*dz/dx = 0 solving for dz/dx gives you: dz/dx = -(Fx/Fz) similarly, dz/dy = -(Fy/Fz) now, use your first order conditions to get the critical point. moved here Link to comment Share on other sites More sharing options...
Guest Posted January 19, 2006 Share Posted January 19, 2006 Wouldn't you have to use partial differentiation to solve this? Link to comment Share on other sites More sharing options...
Julius Caro Posted January 19, 2006 Author Share Posted January 19, 2006 (edited) Yup, it's partial differentation with those formulas dramz posted. The thing is, I got upset because dz/dy and dz/dx (partial derivatives) contained z in them too... So getting the critical poing was a bit tricky. Both partial derivatives are equal to zero (criteria to get the critical point), so x = y = -z/2 I used that in F(x,y,z) so I would have to simplify to get z. z=1, so x and y = -1/2 To determine if the point was an extremum, I used second order criteria.. with that matrix and the determinant... using x=y=-1/2, z=1.. It was a maximum! Anyway, I was also kinda upset because the "crossed" second order derivatives did not have the exact same formula.. but since x=y in this case, they were the same. A tricky excercise, I should say!! If 'z' was not present in the first derivatives it would be no big deal. Now I'm stuck with another excercise: z = f(x,y) is differentiable, and f(0,0)=0 Directional derivative in (0,0), direction (1,2) = 1 Directional derivative in (0,0), direction (1,-1) = -1 a) Calculate Grad(0,0) b) Calculate the plane tangent to the surface in (0,0,0) I know how to do the first part. Directions are (sqrt(5)/5,2sqrt(5)/5) and (sqrt(2)/2,-sqrt(2)/2)... Gradient is (fx(x,y),fy(x,y)) ... So I get a lineal system of two equations, the unknown are fx(0,0) and fy(0,0).. not that hard. But I have no idea on how to get the tanget plane on that point. The formula I have says that I have to do F(x,y,z)=0 (beeing z=f(x,y)), and then the plane is Fx(x-x0)+Fy(y-y0)+Fz(z-z0) I have nooo idea :blush: Edit Update I think... that if I do F(x,y,z) = f(x,y) - z = 0 Maybe I can get something!! Edited January 19, 2006 by Julius Caro Link to comment Share on other sites More sharing options...
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