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This example shows how to solve a scalar minimization problem with nonlinear inequality constraints. The problem is to find x that solves

subject to the constraints

x₁x₂ – x₁ – x₂ ≤ –1.5,
x₁x₂ ≥ –10.

Because neither of the constraints is linear, you cannot pass the constraints to fmincon at the command line. Instead you can create a second file, confun.m, that returns the value at both constraints at the current x in a vector c. The constrained optimizer, fmincon, is then invoked. Because fmincon expects the constraints to be written in the form c(x) ≤ 0, you must rewrite your constraints in the form

Step 1: Write a file objfun.m for the objective function.

function f = objfun(x)
f = exp(x(1))*(4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);

Step 2: Write a file confun.m for the constraints.

function [c, ceq] = confun(x)
% Nonlinear inequality constraints
c = [1.5 + x(1)*x(2) - x(1) - x(2);     
     -x(1)*x(2) - 10];
% Nonlinear equality constraints
ceq = [];

Step 3: Invoke constrained optimization routine.

x0 = [-1,1];     % Make a starting guess at the solution
options = optimoptions(@fmincon,'Algorithm','sqp');
[x,fval] = ... 
fmincon(@objfun,x0,[],[],[],[],[],[],@confun,options);

fmincon produces the solution x with function value fval:

x,fval
x = 
   -9.5474  1.0474 
fval =
    0.0236

You can evaluate the constraints at the solution by entering

[c,ceq] = confun(x)

This returns numbers close to zero, such as

c =

   1.0e-14 *

   -0.6661
    0.7105


ceq =

     []

Note that both constraint values are, to within a small tolerance, less than or equal to 0; that is, x satisfies c(x) ≤ 0.