SOLUTION: An artist is painting a supply of small paintings to sell at an arts festival. He can paint three landscapes per hour and two seascapes. He can frame five paintings per hour. He ha
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Question 93521This question is from textbook Finite mathematics
: An artist is painting a supply of small paintings to sell at an arts festival. He can paint three landscapes per hour and two seascapes. He can frame five paintings per hour. He has 50 hours available for painting and 25 hours for framing. How many of each type of painting should he paint and frame in order to maximize the total value of the paintings. He receives $25 each for the landscapes and $30 each for the seascapes. This question is from textbook Finite mathematics
You can put this solution on YOUR website! An artist is painting a supply of small paintings to sell at an arts festival. He can paint three landscapes per hour and two seascapes. He can frame five paintings per hour. He has 50 hours available for painting and 25 hours for framing. How many of each type of painting should he paint and frame in order to maximize the total value of the paintings. He receives $25 each for the landscapes and $30 each for the seascapes.
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Let x = no. of landscapes; y = no. of seascapes
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The painting equation:
x/3 + y/2 =< 50
Get rid of the denominators, multiply by 6 and you have:
2x + 3y =< 300
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The framing equation:
(x + y)/5 =< 25
Get rid of the denominators, multiply equation by 5
x + y =< 125
y = y = 125 - x; use this for substitution in the Painting equation
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By solving this system of equations we should be able to find the solution.
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Substitute for y and find x:
2x + 3y = 300
2x + 3(125-x) = 300
2x + 375 - 3x = 300
2x - 3x = 300 - 375
-x = -75
x = 75 landscapes
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y = 125 - 75 = 50 seascapes
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Revenue from this:
landscape$ + seascape$
25(75) + 30(50) =
1875 + 1500 = $3375
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If we graph this, it is apparent that x=75, y=50 would give us max revenue
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Graph y = 125-x and y = 100 - (2/3)x
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Area of feasibility, at or below the lowest graphing line.
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