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- Ian HLv 74 weeks ago
Quantity demanded in units per period, (for example daily sales) represented by Q

The variable representing price is p, so the demand function is

Sales Q = 150 – 10p, where p is price

Revenue R is price per item * number sold

R = p*Q = 150p – 10p^2

https://www.wolframalpha.com/input/?i=R+%3D+150p+%...

a) To find maximum R, differentiate and set that result equal to zero.

dR/dp = 150 – 20p = 0

Revenue R is at a max when price p = 7.5

Sales Q when total revenue is at its max is given by Q = 150 – 75 = 75

b) The question repeats “find price at which total revenue is a max”

It seems likely that this should have asked for that max revenue.

If so, R max = 150*7.5 – 10*7.5^2 = 562.5

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