Question 94522
A small company produces both doll houses and sets of doll furniture.
Let the number of houses be "h" and the number of furnitures be "f"
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The doll houses take 3 hours of labor to produce, and the furniture sets take 8 hours.  
The labor available is limited to 400 hours per week, and the total production capacity is 100 items per week. 
Labor inequality: 3h+8f<=400
Production inequality: h+f<=100
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Existing orders require that at least 20 doll houses and 10 sets of furniture be produced per week.
h>=20
f>=10
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Write a system of inequalities representing this situation, where h is the number of doll houses and f is the number of furniture sets.
Solve each inequality for f as follows:
Labor: f(h) = -(3/8)h+50
Production: f(h)<=-h+100
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Then graph the system of inequalities
Label the horizontal axis as "h"; Label the vertical axis as "f"
Draw a vertical line at h=20
Draw a horizontal line at f=10
Comment: Remember solutions are restricted to f>=10 and h>=20
{{{graph(400,300,-5,150,-5,60,10,-(3/8)x+50,-x+100)}}}
Comment: I can't shade the proper inequality areas on this site.
Hopefully you can do that.
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Cheers,
Stan H.