Question 884240
Instead of x[1] and x[2], I'll use x and y.
First find the region and the vertices.
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{{{5x+3y=15}}}
{{{graph(300,300,-2,10,-2,10,(15-5x)/3)}}}
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Now add {{{x+y=6}}}
{{{graph(300,300,-2,10,-2,10,(15-5x)/3,6-x)}}}
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So the feasible region looks like this,
{{{drawing(300,300,-1,8,-1,8,grid(1),circle(0,5,0.2),circle(0,6,0.2),circle(3,0,0.2),circle(6,0,0.2),blue(line(0,5,0,6)),blue(line(0,6,6,0)),blue(line(6,0,3,0)),blue(line(3,0,0,5)))}}}
The extreme values occur at the vertices. 
(0,5) : {{{z=0+5=5}}}
(0,6) : {{{z=0+6=6}}}
(6,0) : {{{z=6+0=6}}}
(3,0) : {{{z=3+0=3}}}
The maximum value occurs along the line {{{x+y=6}}}