% definiramo funkcijo za risanje
f = @(x,y) exp(x.^2 - y.^2);

figure
[xx,yy] = ndgrid(linspace(-1,1),linspace(-1,1));
contour(xx,yy,f(xx,yy),50,'LineWidth',1)
hold on

fimplicit(@(x,y) (x-1/3).^2 + (y-1/3).^2 - (1/3)^2,'r','LineWidth',1);
axis equal


% risemo problem iskanja vezanega ekstrema
f = @(x) exp(x(1)^2 - x(2)^2);



% % ce bi zeleli iskati minimum brez dodatnih omejitev
% xmin = fminsearch(f,[0;0])
% 
% % ce bi zeleli iskati minimum na obmocju [-1,1]^2
% xmin = fmincon(f,[.6;.6],[],[],[],[],[-1 -1],[1 1])

xmin = fmincon(f,[.6;.6],[],[],[],[],[-1 -1],[1 1],@circlecon)
xmax = fmincon(@(x)-f(x),[.0;.0],[],[],[],[],[-1 -1],[1 1],@circlecon)

plot(xmin(1),xmin(2),'ob','MarkerEdgeColor','k', 'MarkerFaceColor',[0 0 1],'MarkerSize',5)
plot(xmax(1),xmax(2),'or','MarkerEdgeColor','k', 'MarkerFaceColor',[1 0 0],'MarkerSize',5)


% nelinearens pogoj: kroznica
function [cineq,ceq] = circlecon(x)
    cineq = [];
    ceq = (x(1)-1/3)^2 + (x(2)-1/3)^2 - (1/3)^2;
end