Finding x and y values of minimum z in 2-variable function.

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Alex
Alex 2025-2-27
回答： Torsten 2025-2-27
I'm using for loops to find a minimum of z(x,y) as well as the x and y of that value.
With a step size of 1 it finds the right values, but with a step size of 0.1 the y value is 10 when it should be around 2.
Any help would be appreciated.
[x,y] = meshgrid(0:0.1:10, 0:0.1:10);
z = (x.^2).*(y.^3)-1.4*(x.^2).*(y.^2)-3.91*(x.^2).*y+5.78*(x.^2)-...
11*x.*(y.^3)+15.4*x.*(y.^2)+43.01*x.*y-63.58*x+30.25*(y.^3)-...
42.35*(y.^2)-118.278*y+174.845;
min = 100;
for i=1:101
for j=1:101
if z(i,j)<min
min = z(i,j);
xmin = x(i,j);
ymin = y(i,j);
end
end
end
  1 个评论
John D'Errico
John D'Errico 2025-2-27
Note that it is a really bad idea to use a variable neamed min, as then later on when you need to use the function min, MATLAB will not know how to distinguish between the function and variable.

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回答（3 个）

Matt J
Matt J 2025-2-27
编辑：Matt J 2025-2-27
Ran in:
The continuous minimum can't be at (5,2) as you claim. Direct evaluation shows that the value of z at (5.5,10) is lower.
loopingMin(1)
xmin = 5
ymin = 2
minval = 0.0890
loopingMin(0.1)
xmin = 5.5000
ymin = 10
minval = -0.0050
function loopingMin(stepsize)
[x,y] = meshgrid(0:stepsize:10, 0:stepsize:10);
z = (x.^2).*(y.^3)-1.4*(x.^2).*(y.^2)-3.91*(x.^2).*y+5.78*(x.^2)-...
11*x.*(y.^3)+15.4*x.*(y.^2)+43.01*x.*y-63.58*x+30.25*(y.^3)-...
42.35*(y.^2)-118.278*y+174.845;
minval = inf;
for i=1:height(x)
for j=1:width(y)
if z(i,j)<minval
minval = z(i,j);
xmin = x(i,j);
ymin = y(i,j);
end
end
end
xmin,ymin,minval
end
Star Strider
Star Strider 2025-2-27
Ran in:
Is there a particular reason to use that sort of iteration? MATLAB has a number of optimization funcitons you can use, with fminsearch being a part of core MATLAB (no Toolboxes required) or one of the Optimization Toolbox functions, such as fsolve
That aside, I keep getting different results betweeen runs and between functions —
z = @(x,y) (x.^2).*(y.^3)-1.4*(x.^2).*(y.^2)-3.91*(x.^2).*y+5.78*(x.^2)-...
11*x.*(y.^3)+15.4*x.*(y.^2)+43.01*x.*y-63.58*x+30.25*(y.^3)-...
42.35*(y.^2)-118.278*y+174.845;
B0 = randn(2,1)
B0 = 2×1
-0.6770 0.9243
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[B,fv] = fminsearch(@(b) norm(z(b(1),b(2))), B0)
B = 2×1
0.0204 1.6972
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
fv = 3.3765e-11
zval = z(B(1),B(2))
zval = -3.3765e-11
[B,fv] = fsolve(@(b) norm(z(b(1),b(2))), B0)
Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt algorithm instead.
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
B = 2×1
-0.6004 1.6975
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
fv = 8.6173e-07
zval = z(B(1),B(2))
zval = 8.6173e-07
[B,fv] = fminunc(@(b) norm(z(b(1),b(2))), B0)
Local minimum possible. fminunc stopped because it cannot decrease the objective function along the current search direction.
B = 2×1
-0.5319 1.7025
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
fv = 4.0937e-09
[X,Y] = ndgrid(linspace(min(B)-0.1*abs(min(B)), max(B)+0.1*abs(max(B)), 250));
figure
surfc(X, Y, z(X,Y), EdgeColor='interp', FaceAlpha=0.1)
hold on
stem3(B(1), B(2), z(B(1),B(2))+5, 'vr', MarkerFaceColor='r')
hold off
grid
zlim([min([zval zlim]) max(zlim)])
xlabel('X')
ylabel('Y')
zlabel('Z')
colormap(turbo)
view(60,30)
grid on
.
Torsten
Torsten 2025-2-27
Ran in:
z = @(x,y)(x.^2).*(y.^3)-1.4*(x.^2).*(y.^2)-3.91*(x.^2).*y+5.78*(x.^2)-...
11*x.*(y.^3)+15.4*x.*(y.^2)+43.01*x.*y-63.58*x+30.25*(y.^3)-...
42.35*(y.^2)-118.278*y+174.845;
sol = fmincon(@(u)z(u(1),u(2)),[1 1],[],[],[],[],[0 0],[10 10])
Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
sol = 1×2
5.5000 9.9840
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
z(sol(1),sol(2))
ans = -0.0050

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