How to define multiple constraints for fmincon?

Hi, I'm trying to minimize the formula below. I've done most of the code except I can't quite figure out how to account for the constraints. I tried at the end of the code but I think this is for nonlinear constraints? And mine are linear. Also not sure how to account for the bounds.

Anyone know how I can do this? Thanks!

%FORMULA: SC= 3SA + 4SS + 6LA + 4LS

%BOUNDS:
%SA>= 100 locks
%SS>=100 locks
%LA>=100 locks
%LS>=100 locks

%CONSTRAINTS: 
%SA + LA = 2000
%SS + LS = 1500
%LA + LS <= 1500
%SA + SS <= 2500
%All variables need to be positive & integers


%load guesses into array
x0=[SA SS LA LS];


%call solver to minimize objective function given the constraint
xopt= fmincon(@objective, x0, [],[],[],[],[],[],@constraint,[])

%retrieve optimized shipping cost
SC_Opt=ship_cost(xopt)

%double check constraints


%function to calculate FendPac total shipping cost
function  ship_cost= ship_cost(x)
    SA= x(1);
    SS= x(2);
    LA = x(3);
    LS = x(4);
    ship_cost=3*SA + 4*SS + 6*LA + 4*LS;
end 


%function to calculate shipped locks to Austin
function to_atx = to_atx(x)
    SA= x(1);
    LA= x(2);
    to_atx= LA + SA
end 

%function to calculate shipped locks to San Antonio

function to_satx = to_satx(x)
    SS= x(1);
    LS= x(2);
    to_satx= LS + SS
end 

%function to calculate shipped locks from Los Angeles

function from_LA = from_LA(x)
    LS= x(1);
    LA= x(2);
    from_LA= LS + LA
end 

%function to calculate shipped locks from San Francisco

function from_SF = from_SF(x)
    SS= x(1);
    SA= x(2);
    from_SF= SS + SA
end

%objective function for optimization
function obj=objective(x)
    obj=ship_cost(x);
end


%CONSTRAINTS
function [c,ceq]= constraint1(x)
    c(1)= 2000- to_atx(x);
    c(2)= 1500- to_satx(x);
    c(3)=from_LA(x) - 1500;
    c(4)= from_SF(x) - 2500
   
    ceq1=[];
end 

[Walter Roberson]

function to_atx = to_atx(x)
    SA= x(1);
    LA= x(2);
    to_atx= LA + SA
end 

No, LA is x(3) not x(2) .

Answer 1

[Walter Roberson]

Aeq = [1 0 1 0; 0 1 0 1];
beq = [2000; 2000];
A = [0 0 1 1; 1 1 0 0];
b = [1500; 2500]
lb = [100, 100, 100, 100];
ub = [1900, 1900, 1900, 1900];  %min 100, sum = 2000, so max = 1900

xopt = fmincon(@objective, x0, A, b, Aeq, beq, lb, ub)