Question

79. Revenue. The revenue (in dollars) from the sale of x infant car seats is given by R1x2 = 60x - 0.025x2 0 ... X ... 2,400 (A) Find the average change in revenue if production is changed from 1,000 car seats to 1,050 car seats. (B) Use the four-step process to find R′1x2. (C) Find the revenue and the instantaneous rate of change of revenue at a production level of 1,000 car seats, and write a brief verbal interpretation of these results.

Answer 1

Question:

59 Revenue. The revenue (in dollars) from the sale of x infant car seats for infants is given by

R(x) =60·x - 0.025·x² 0 ≤ x ≤ 2400

(A) Find the average change in revenue if production is changed from 1,000 car seats to 1,050 car seats.

(B) Use the four-step process to find

(C) Find the revenue and the instantaneous rate of change of revenue at a production level of 1,000 car seats, and write a brief verbal interpretation of these results

Answer:

(A) $437.5

(B) R'(x) = 60 - 0.05·x

(C) $10

Step-by-step explanation:

(A) Here we have the change in revenue when production is increased from 1,000 car seats to 1,050 car seats is given as follows;

R(1050) - R(1000) = 60×1050 - 0.025×1050² - (60×1000 - 0.025×1000²)

R(1050) - R(1000) = 63000 - 27562.5 - 35000 = $437.5

(B) R'(x) is given by

Step 1. R(x + h) = 60·(x+h) - 0.025·(x+h)²

Step 2  R(x + h) - R(x) = 60·(x+h) - 0.025·(x+h)² -  60·x - 0.025·x² = $-\frac{h^2+(2x-2400)h}{40}$

Step 3

$\frac{R(x+h)- R(x)}{h} = -\frac{h^2+(2x-2400)h}{40 \times h} = -\frac{h +2x - 2400}{40}$

Step 4

$\lim_{h \to 0} \frac{R(x+h)- R(x)}{h} = \lim_{h \to 0} -\frac{ h +2x - 2400}{40}$

∴  R'(x) = 60 - 0.05·x

(c) The revenue at a production level of 1000 car seats is

R(1000) = (60×1000 - 0.025×1000²) = $35,000

The instantaneous rate of change of revenue is given as follows

R'(x) = 60 - 0.05·x = 60 - 0.05×1000 = $10.