8/10/2014 A computer manufacturer determines that it can sell 4000 machines at $750, and that for each $25 that the price is increased 20 fewer computers are sold. How much the manufacturer should charge to maximize his revenue?



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8/10/2014 Solution. If n represents the number of $25 price increases, the Revenue function, R(n) = (4000 – 20n)(750 + 25n)

= 500(6000 + 170n – n2)

= -500n2 + 85000n + 3000000, which is maximized for

n = -b/2a = = -85000/(2-500) = 85

So, the manufacturer should set the price at ($750 + $25∙85) = $2125 at which price $6612500 shall be made by selling 2300 computers.