1.

A = [1 2 3; -1 3 7; 9 0 3]
b = [1; 4; 7]

x = A \ b

L, U, P = lu(A)
y = L \ b[P]
x = U \ y

2.

using TyOptimization
c = [-40, -30]
A = [3 2]
b = [1000]
lb = [0, 0]
x, fval, = linprog(c, A, b, [], [], lb)
println("最优解为", x)
println("利润为", -fval)

f = [-40, -30]
intcon = [1, 2]
A = [3 2]
b = [1000]
lb = [0, 0]
x, fval, = intlinprog(f, intcon, A, b, [], [], lb)
println("最优解为", x)
println("利润为", -fval)