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 2020-09-10 10:55:54 +0100 commented answer How to find the symbolic maximum of a 3D function? Thanks for the solution and explanation. With resolution 400 or higher the function looks like a razor blade, with the maximum infinite fare away. At a high resolution of 1600, KDE shows me "An application has crashed". So there is no local maximum and that explaines why SageMath and Mathematica could not find a local maximum. That the derivatives are zero at (0, 0) is trivial, because that means switching the polarity of the symmetric machine, which does not affect the output power/function f. So f is point symmetrical to (0,0). 2020-09-09 21:22:47 +0100 commented answer How to find the symbolic maximum of a 3D function? Thanks for the new answers and working on it! Setting a=b=1 is only a workaround for plotting. a and b are machine parameters and independet from another. 2020-09-08 11:09:56 +0100 commented answer How to find the symbolic maximum of a 3D function? Thanks for the answer, but at (0, 0) is a saddle point, with value zero. In the 3D plot with a=1, b=1, i clearly see one maximum at appromimately (4, 3). I need that maximum, as function of a and b. The reason why i need it, is that f(x,y) is the output power of a machine and i want to know the maximum power point (MPP). http://www.true-random.com/files/unsorted/P_out_a_eq_b_eq_1_2020-09-06a.png (Plot of f(x,y) starting from (0,0)) http://www.true-random.com/files/unsorted/P_out_a_eq_b_eq_1_2020-09-06b.png (Plot of max_symbolic(0, f(x,y)) starting from (0,0)) 2020-09-08 02:01:36 +0100 asked a question How to find the symbolic maximum of a 3D function? I have a function of two real variables, f(x,y) = a * (x * y - y^2) * exp( -b * x^2) * exp(b * x * y ) with the real parameters a and b, and five restrictions: x>0; y>0; a>0; b>0; y<=x (y equal to or less than x) In the 3d_plot, with a=b=1, i see there is only one local maximum, which is also the global maxium. How can i get the formula, for x and y coordinate of the maximum, as function of the parameters a and b? I tried assume() for the restrictions, but that does not work and is made for the other way, to tell what is already there, not what should be changed. To get rid of the negative function values, i used r(x,y) = max_symbolic(0, f(x,y)) and that works, but that is only a workaround for the not implemented restrictions.