Use the simplex method to maximize the following:

Maximize f=19x1+3x2+20x3\displaystyle {f}={19}{x}_{{1}}+{3}{x}_{{2}}+{20}{x}_{{3}} subject to
2x1+x2+x340\displaystyle {2}{x}_{{1}}+{x}_{{2}}+{x}_{{3}}\le{40}
x1+2x260\displaystyle {x}_{{1}}+{2}{x}_{{2}}\le{60}
x2+3x320\displaystyle {x}_{{2}}+{3}{x}_{{3}}\le{20}

x10,x20,x30\displaystyle {x}_{{1}}\ge{0},{x}_{{2}}\ge{0},{x}_{{3}}\ge{0}

If no solutions exist enter DNE in all answerboxes.

x1=\displaystyle {x}_{{1}}=

x2=\displaystyle {x}_{{2}}=

x3=\displaystyle {x}_{{3}}=

f=\displaystyle {f}=