Question 330597
Previously answered.
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Let A be the number of Model A's and B be the number of model B's.
Time Available
1.{{{2A+3B<=34}}}
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Money available
{{{25A+30B<=350}}}
2.{{{5A+6B<=70}}}
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Let A be represented by the x axis and B by the y axis.
Plot line 1, {{{B=(34-2A)/3}}} and plot line 2,{{{B=(70-5A)/6}}} and find the intersection point when 
{{{(34-2A)/3=(70-5A)/6}}}
{{{68-4A=70-5A}}}
{{{A=2}}} and {{{B=10}}}
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{{{drawing(300,300,-2,15,-2,15,grid(1),circle(2,10,0.4),circle(14,0,.4),circle(0,34/3,.4),graph(300,300,-2,15,-2,15,(34-2x)/3,(70-5x)/6),blue(line(0,34/3,2,10)),blue(line(2,10,14,0)),blue(line(0.1,0.1,0.1,34/3)),blue(line(0.1,0.1,14,0.1)))}}}
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The company could assemble any combination of A and B within the blue box with vertices:(0,0),(0,34/3),(2,10),(14,0).
As an example, {{{A=4}}}, {{{B=6}}} would work. However they would have hours and money left over. 
If they want to use exactly $350 and 34 hours, then the only solution is {{{A=2}}}, {{{B=10}}}