Calculus : Price Function

Question

For the price function

p(x) = 10 - √3x

Where p is in dollars and x is the number sold in hundreds (0 ≤ x ≤ 30), use R(12) and MR(12) to approximate the revenue when 1260 items are sold. Then interpret the marginal revenue value. At this level of sales the revenue increases by about $_____ per additional hundred items. How much will the price need to be changed to have this level of sales?

Answer 1

P = 10- sqr3x

Revenue = R = Px = 10x -xsqr3x = 10x -(sqr3)(x)^(3/2)

MR = R' = 10-(3/2)(sqr3x)

R(12) = 10(12) - 12sqr36 = 120-72 =48

MR(12) = 10-(3/2)(sqr36) = 1

If output increases by 1, revenue increases by 1

R(13) = 10(13) -13(sqr39) = about 48.8, an increase of 0.8, almost 1

If output increases by 0.6, revenue would be estimated to increase by 0.6

actual R(12.6)= 10(12.6) - 12.6sqr(37.8) = 126- 48.5

the estimate is off by only 0.1

MR(12.6) = 10-(3/2)(sqr37.8) = about 0.78 increase for 100 more output

p = 10-sqr3(12.6) = 10-sqr37.8 = 3.85