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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

*x*_{1}*x*_{2} – *x*_{1} – *x*_{2} ≤
–1.5,*x*_{1}*x*_{2} ≥
–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

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);

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 = [];

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.

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