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the price p and the quantity x sold of a certain product obey the demand equation below

x= -6p+132,   0_<p_<30

Express the revenue R as a function of x. What is the revenue if18 units are sold. what quantity x maximizes revenue. what is the maximum revenue. what price sold the company charge to maximize revenue. what price should the company charge to earn at least 630 in revenue.

Apr 1st, 2015

OK, Revenue = Price * Sales:

= x*p

= (-6p + 132)p

R(p) = -6p^2 + 132p

Or as a function of x (why??):

p = 22 - x/6

R(x) = 22x - x^2 / 6

If 18 units are sold then:

x = 18

R(x) = 22x - x^2 / 6

= 342

Optimal price when R(p) is maximised, so solve R'(p) = 0:

R'(p) = -12p + 132

0 = -12p + 132

p = 11

Revenue at p = 11:

R(p=11) = -6(11^2) + 132(11)

R = 726

For Revenue to be at least 630:

R(p) >= 630

-6p^2 + 132p >=630

Set -6p^2 + 132p = 630, then

p^2 - 22p + 105 = 0

(p - 7)(p - 15) = 0

So 7<= p <= 15 earns revenue >= 630.

Apr 1st, 2015

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Apr 1st, 2015
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Apr 1st, 2015
Dec 3rd, 2016
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