The third constraint can be re-written as −x2≥3. The solution space satisfying constraints is shown shaded in the figure.
A(−3,−3) and B(10,−3) value of the objective functions Z=−x1+4x2 at these vertices are
Z(A)=3−12=−9, Z(B)=−10−12=−22
Thus the maximum value of Z occurs at A. Hence the solution to the problem is
x1=−2,x2−3;Zmax=−9