If i have a long function with multiple variables, How can i find its absolute max and absolute min ?

i have this function---> f=[(268/((3*x+18)^2+(3*y+7)^2+(18+7)))-(2*(268)/((3*x-21)^2+(3*y+21)^2+(21+21)))-(3*(268)/((3*x+18)^2+(3*y-21)^2+(18+21)))+(4*(268)/((3*x-21)^2+(3*y-7)^2+(21+7))) ] I'm trying to find a away to calculate the absolute max and the absolute min. Everything i tried wont work, because my function's format does not fit their criteria.

Please help! Jerry,

Answer 1

[Star Strider]

I’m not certain what you’ve tried, or the result you want. I used the Symbolic Math Toolbox to simplify your equation and turn it into a function (with simplify and matlabFunction), then used fminsearch (experimenting with several initial parameter estimates) and plotted it over an inclusive range of ‘x’ and ‘y’:

   f = @(x,y)2.68e2./((x.*3.0+1.8e1).^2+(y.*3.0+7.0).^2+2.5e1)+1.072e3./((x.*3.0-2.1e1).^2+(y.*3.0-7.0).^2+2.8e1)-8.04e2./((x.*3.0+1.8e1).^2+(y.*3.0-2.1e1).^2+3.9e1)-5.36e2./((x.*3.0-2.1e1).^2+(y.*3.0+2.1e1).^2+4.2e1);

f_min = fminsearch(@(b)  f(b(1),b(2)),  [5; 5]);
f_max = fminsearch(@(b) -f(b(1),b(2)),  [5; 5]);

x = linspace(-15, 15, 50);                                    % Define ‘x’ Range
y = linspace(-15, 15, 50);                                    % Define ‘y’ Range
[X,Y] = meshgrid(x,y);

figure(1)
meshc(X, Y, f(X,Y))
hold on
plot3(f_min(1), f_min(2), f(f_min(1), f_min(2)), 'vr', 'MarkerFaceColor','r')
plot3(f_max(1), f_max(2), f(f_max(1), f_max(2)), '^r', 'MarkerFaceColor','r')
hold off
grid on
view([-160  25])

Experiment with other initial parameter estimates for the function. I doubt you will arrive at a different ‘global’ maximum and minimum.

EDIT — Added plot.