Pierce Manufacturing determines that the daily​ revenue, in​ dollars, from the sale of x lawn chairs is ​R(x)equals0.006 x cubed plus 0.02 x squared plus 0.5 x. ​Currently, Pierce sells 50 lawn chairs daily. ​a) What is the current daily​ revenue? ​b) How much would revenue increase if 54 lawn chairs were sold each​ day? ​c) What is the marginal revenue when 50 lawn chairs are sold​ daily? ​d) Use the answer from part​ (c) to estimate ​R(51​), ​R(52​), and ​R(53​).

Respuesta :

Answer:

a) $825

b) $205.10

c) $47.5

d) R(51) = 872.5, R(52) = 920, R(53) = 967.5

Step-by-step explanation:

We are given the following information in the question:

[tex]R(x) = 0.006x^3 + 0.02x^2 + 0.5x[/tex]

where x is the number of lawn chairs.

a) Current daily​ revenue

[tex]R(50) = 0.006(50)^3 + 0.02(50)^2 + 0.5(50) = 825[/tex]

Thus, $825 is the current daily​ revenue when Pierce sells 50 lawn chairs daily.

b) Revenue increase if 54 lawn chairs were sold each​ day

[tex]R(54)-R(50)\\= 0.006(54)^3 + 0.02(54)^2 + 0.5(54) - 825\\= 205.10[/tex]

Thus, revenue increase $205.10 if 54 lawn chairs were sold each​ day.

c) Marginal revenue

[tex]R'(x) = \displaystyle\frac{dR}{dx} = \frac{d(0.006x^3 + 0.02x^2 + 0.5x)}{dx}\\\\R'(x) = 0.018x^2 + 0.04x+0.5\\R'(50) = 0.018(50)^2 + 0.04(50)+0.5 = 47.5[/tex]

Thus, $47.5 is the marginal revenue when 50 lawn chairs are sold​ daily.

d) R(51​), ​R(52​), and ​R(53​)

[tex]R(51) = R(50) + 1R'(50) = 825 + 1(47.5) = 872.5\\R(52) = R(50) + 2R'(50) = 825 + 2(47.5) = 920\\R(53) = R(50) + 3R'(50) = 825 + 3(47.5) = 967.5[/tex]