Question 466150
The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p, in dollars, that it charges. If the revenue R, in dollars, is 
R(p)=-1/2p²+1900p

• (a) At what prices p is revenue zero?
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Solve: (-1/2)p^2+1900p = 0
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Factor:
p[(-1/2)p + 1900] = 0
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p = 0 or p = 3800
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• (b) For what range of prices will revenue exceed $1,200,000?
Solve: (-1/2)p^2+1900p > 1,200,000
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(-1/2)p^2+1900p - 1200000 > 0
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(-(1/2))[p^2-3800p+ 2400000 > 0
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p^2 - 3800p + 2400000 < 0
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Graph:
{{{graph(400,300,-10000,10000,-2400000, 2400000, x^2-3800x+2400000)}}}
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Solution: 800 < p < 3000 

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Cheers,
Stan H.