Question 1042559
Calculators are sold to students for 1000 dollars each. Five hundred students are willing to buy them at that price. For every 20 dollar increase in price, there are 30 fewer students willing to buy the calculator.
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selling price:: 1000+20x
# of student buyers:: 500-30x
Revenue = price*# of buyers = (1000+20x)(500-30x)
R(x) = -600x^2 - 20000x + 500000
a. What selling price will produce the maximum revenue and what will be the maximum revenue?
R'(x) = -1200x - 20000
Solve:: R'(x) = 0

-1200x - 20000 = 0
-12x  = 200
x = -16 2/3
Ans: 1000+20(-16 2/3) = $666.80
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b. Show and illustrate the table of values, equation and graph for the revenue function.
{{{graph(400,400,-50,50,-10000,500000,-600x^2 - 20000x + 500000)}}}
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c.State the domain, range, and x-intercepts of the graph.
Range:: [0,666666.67]
X-intercepts:: [-50,16 2/3] 
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Cheers,
Stan H.
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