Question 341432
Determine the feasible region by plotting the constraint equations.
{{{drawing(300,300,-2,10,-2,10,red(line(7,-10,7,10)),grid(1),graph(300,300,-2,10,-2,10,(40-5x)/2,(48-3x)/6,2x-3))}}}
Find the intersection points.
Between the x-axis and {{{y=2x-3}}}
{{{2x-3=0}}}
{{{2x=3}}}
{{{x=3/2}}}
({{{3/2}}},{{{0}}})
The other intersection point is ({{{7}}},{{{0}}}).
Find the intersection points of {{{5x+2y=40}}} and {{{x=7}}}
{{{35+2y=40}}}
{{{2y=5}}}
{{{y=5/2}}}
({{{7}}},{{{5/2}}})

{{{drawing(300,300,-2,8,-2,8,
circle(3/2,0,0.2),
circle(7,5/2,0.2),
circle(7,0,0.2),
red(line(7,-10,7,10)),grid(1),graph(300,300,-2,8,-2,8,(40-5x)/2,(48-3x)/6,2x-3))}}}

Find the intersection point between
1. {{{5x+2y=40}}} 
2. {{{3x+6y=48}}}
From eq. 2,
{{{x+2y=16}}}
{{{2y=16-x}}}
Substitute into eq. 1,
{{{5x+16-x=40}}}
{{{4x=24}}}
{{{x=6}}}
Then from eq. 2,
{{{2y=16-6}}}
{{{y=5}}}
({{{6}}},{{{5}}})
Finally find the intersection point between {{{y=2x-3}}} and {{{3x+6y=48}}}
{{{2y=4x-6}}}
Substitute,
{{{x+2y=16}}}
{{{x+4x-6=16}}}
{{{5x=22}}}
{{{x=22/5}}}
Then
{{{y=44/5-15/5}}}
{{{y=29/5}}}

({{{22/5}}},{{{29/5}}})
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{{{drawing(300,300,-2,10,-2,10,
circle(3/2,0,0.2),
circle(7,5/2,0.2),
circle(7,0,0.2),
circle(22/5,29/5,0.2),
circle(6,5,0.2),
red(line(7,-10,7,10)),grid(1),graph(300,300,-2,10,-2,10,(40-5x)/2,(48-3x)/6,2x-3))}}}
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The maximum and minimum of the function occurs at one of these vertices.
({{{3/2}}},{{{0}}}) :{{{F=5x+3y=5(3/2)=15/2}}}
({{{7}}},{{{0}}}):{{{F=5x+3y=5(7)=15/2}}}

({{{7}}},{{{5/2}}}):{{{F=5x+3y=5(7)+3(5/2)=70/2+15/2=85/2}}}

({{{6}}},{{{5}}}):{{{F=5x+3y=5(6)+3(5)=45}}}

({{{22/5}}},{{{29/5}}}):{{{F=5x+3y=5(22/5)+3(29/5)=110/5+87/5=197/5}}}
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The maximum value is {{{45}}} and occurs at ({{{6}}},{{{5}}}).