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Question

Each month a store owner can spend at most 100,000onPCslaptops.APCcoststhestoreowner 1000 and a laptop costs him 1500.EachPCissoldforaprofitof 400 while laptop is sold for a profit of 700.Thestoreownerestimatesthatatleast15PCsbutnomorethan80$ are sold each month. He also estimates that the number of laptops sold is at most half the PC's. How many PC's and how many laptops should be sold in order to maximize the profit?

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Solution

Let x and y be the numbers of PC's and laptops respectively that should be sold.
Profit =400x+700y to maximize
Constraints
15x80 least 15 PC's but no more than 80 are sold each month
y(1/2)x
1000x+1500y100,000 "store owner can spend at most 100,000$ on PC's and laptops"
⎪ ⎪ ⎪⎪ ⎪ ⎪15x80y0y(1/2)x1000x+1500y100,000
The solution set of the system of inequalities above and the vertices of the feasible solution set obtained are shown.
Vertices:
A at intersection of x=15 and y=0 (x-axis) coordinates of A:(15,0)
B at intersection of x=15 and y=(1/2)x coordinates of B:(15,7.5)
C at intersection of y=(1/2)x and 1000x+1500y=100000 coordinates of C:(57.14,28.57)
D at intersection of 1000x+1500y=100000 and x=80 (y-axis) coordinates of D:(80,13,3)
Evaluate the profit at each vertex
A(15,0),P=400×15+700×0=6000
B(15,7.5).P=400×15+700×7.5=11250
C(57.14,28.57),P(400×57.14+700×28.57=42855
D(80,13.3),P=400×80+700×13.3=41310
The profit is maximum for x=57.14 and y=28.57 but these cannot be accepted as solutions because x and y are numbers of PC's and ;laptops and must be integers. We need to select the nearest integers to x=57.14 and y=28.57 that are satisfy all constraints and give a maximum profit
x=57 and y=29 do not satisfy all constraints
x=57 and y=28 satisfy all constraints
Profit=500×57+700×28=42400, which is maximum
1032577_849539_ans_39c90799bf2b4748baedfae7f4d1ab9e.PNG

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