Question 322796
Solve the following Linear Programming problem using the corner point method. 
{{{P(x,y)=3x + 5y}}}


{{{drawing(300,300,-2,12,-2,12,grid(1),circle(0,0,.3),circle(12,0,.3),circle(0,10,0.3),circle(4,8,0.3),
blue(line(4,8,0,10)),
blue(line(4,8,12,0)),
blue(line(0,10,0,0)),
blue(line(12,0,0,0)),
blue(line(0.1,0.1,0.1,10)),
blue(line(0.1,0.1,12,0.1)),
green(line(-10,22,4,8)),
green(line(15,-3,12,0)),
red(line(-10,15,0,10)),
red(line(4,8,16,2)),
graph(300,300,-2,12,-2,12,0))}}}
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The feasible region (shown by the blue polygon) is formed from the two lines and the two axes. The intersection of the lines with the axes give two points, the intersection of the axes (0,0) gives a third point. 
You also need to find the intersection of the two lines to get the fourth point in the feasible region:
{{{y=12-x}}} and
{{{y=(20-x)/2}}}
which occurs at (4,8).
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c)The maximum and minimum for P(x,y) occurs at one of the vertices.
Calculate the value at each of them,
({{{0}}},{{{0}}}):{{{P=3x+5y=0}}}
({{{12}}},{{{0}}}):{{{P=3x+5y=3(12)=36}}}
({{{0}}},{{{10}}}):{{{P=3x+5y=5(10)=50}}}
({{{4}}},{{{8}}}):{{{P=3x+5y=3(4)+5(8)=52}}}
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d)The maximum profit of {{{52}}} occurs when {{{x=4}}} and {{{y=8}}}.