how to find the minimum value of the function of two variables with contour plot?

How to find minimum value of the function below? I was able to make the contour plot but unable to get the min value using the fminsearch. It kept giving error message when I tried to get the min value. Below is the function and the contour I tried:

x = linspace(0.5,6);
y = linspace(0.5,6);
[X,Y] = ndgrid(x,y);
Q2 = 4.74/4*(3*exp(-2.054*(X-1.401))-2*exp(-1.027*(X-1.401)));
Q1 = 4.74/4*(3*exp(-2.054*(Y-1.401))-2*exp(-1.027*(Y-1.401)));
Q3 = 4.74/4*(3*exp(-2.054*(X+Y-1.401))-2*exp(-1.027*(X+Y-1.401)));
J2 = 4.74/4*(-exp(-2.054*(X-1.401))+6*exp(-1.027*(X-1.401)));
J1 = 4.74/4*(-exp(-2.054*(Y-1.401))+6*exp(-1.027*(Y-1.401)));
J3 = 4.74/4*(-exp(-2.054*(X+Y-1.401))+6*exp(-1.027*(X+Y-1.401)));

Z = 0.904.*(Q1+Q2+Q3-(sqrt(0.5.*((J2-J1).^2+(J1-J3).^2+(J3-J2).^2))));
contour(X,Y,Z,100,'black');

[Walter Roberson]

There is no unique solution . Your equation is symmetric in X and Y, and the minima occur along the boundary with one of the two variables being 6

[Sophia Sagala]

Is it so there is no unique solution? Yes, it is symmetric and I need to find the saddle point of this contour plot at which the r1=r2 as it is symmetry. But I could not find the right syntax in Matlab. Like in Mathematica the syntax for finding routine minimum is FindMinimum, but I do not have access to this software and am more familiar with Matlab.

[Walter Roberson]

If you make the substituation Y = X in order to find the saddle point, then the minima is at

X = Y = 1.863046875000001 with a function value of about -3.35 . This is not the minima in the area, which has a value of about -4.27 when one of the variables is about 1.401 and the other of the variables is 6.

[Sophia Sagala]

Can you please explain me how did you get the minima at X = 1.863 with that corresponding value? What code did you use? I did substitute Y = X.

Thanks before.

[Walter Roberson]

fminsearch() is enough to get this result. The exact solution is X =

-(1000*ln(3))/1027 - (1000*ln(5))/1027 + (1000*ln((7*exp(1438827/1000000) + 4)*(343*exp(4316481/1000000) + 588*exp(1438827/500000) + 30*exp(1438827/2000000)*sqrt(343*exp(4316481/1000000) + 588*exp(1438827/500000) + 1011*exp(1438827/1000000) + 64)*sqrt(3) + 1686*exp(1438827/1000000) + 64)^(1/3) + 49*exp(1438827/500000) + 56*exp(1438827/1000000) + (343*exp(4316481/1000000) + 588*exp(1438827/500000) + 30*exp(1438827/2000000)*sqrt(343*exp(4316481/1000000) + 588*exp(1438827/500000) + 1011*exp(1438827/1000000) + 64)*sqrt(3) + 1686*exp(1438827/1000000) + 64)^(2/3) + 16))/1027 - (1000*ln(343*exp(4316481/1000000) + 588*exp(1438827/500000) + 30*exp(1438827/2000000)*sqrt(343*exp(4316481/1000000) + 588*exp(1438827/500000) + 1011*exp(1438827/1000000) + 64)*sqrt(3) + 1686*exp(1438827/1000000) + 64))/3081

which I found with Maple.

Answer 1

[KSSV]

[c,h] = contour(X,Y,Z,'showtext','on');
min(h.LevelList)

[Sophia Sagala]

I am sorry the value does not show the right minimum value.