Question 884657
First find the feasible region and vertices.
{{{2x+3y>=6}}}
{{{graph(300,300,-2,8,-2,8,y>=(6-2x)/3)}}}
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{{{3x-2y<=9}}}
{{{graph(300,300,-2,8,-2,8,y>=(3x-9)/2)}}}
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{{{x+5y<=20}}}
{{{graph(300,300,-2,8,-2,8,y<=(20-x)/5)}}}
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{{{drawing(300,300,-2,8,-2,8,grid(1),circle(0,2,0.2),circle(0,4,0.2),circle(3,0,0.2),circle(5,3,0.2),
blue(line(0,2,0,4)),
blue(line(0,4,5,3)),
blue(line(5,3,3,0)),
blue(line(3,0,0,2)))}}}
The vertices here were easy to find graphically but you could have solved for them algebraically also.
The extrema occur at the vertices.
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(0,2):{{{C=4(0)+3(2)=highlight(6)}}}
(0,4):{{{C=4(0)+3(4)=12}}}
(5,3):{{{C=4(5)+3(3)=highlight(29)}}}
(3,0):{{{C=4(3)+3(0)=12}}}