Question

the revenue (in dollars) from the sale of x car seats for infants is given by the following function. r(x)=28x - 0.010x² 0≤x≤2800 (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′(x). (c) find the revenue and the instantaneous rate of change of revenue at a production level of 1,000 car seats, and interpret the results. (a) find the average change in revenue if production is changed from 1,000 car seats to 1,050 car seats. 7.5 (round to one decimal place as needed.) (b) r′(x)=28 - 0.02x (c) r(1000)=18000 r′(1000)=8 interpret these results. choose the correct answer below. a. this means that at a production level of 1,000 car seats, the revenue is r(1000) dollars and is decreasing at a rate of r′(1000) dollars per seat. b. this means that at a production level of 1,000 car seats, the revenue is r(1000) dollars and is increasing at a rate of r′(1000) dollars per seat. c. this means that at a production level of 1,000 car seats, the revenue is r′(1000) dollars and is increasing at a rate of r(1000) dollars per seat.

Explanation:

Step1: Recall average - change formula

The average change of a function $y = f(x)$ from $x = a$ to $x = b$ is $\frac{f(b)-f(a)}{b - a}$. Here, $R(x)=28x - 0.010x^{2}$, $a = 1000$, and $b = 1050$.
$R(1000)=28\times1000-0.010\times1000^{2}=28000 - 10000=18000$.
$R(1050)=28\times1050-0.010\times1050^{2}=29400-11025 = 18375$.
The average change is $\frac{R(1050)-R(1000)}{1050 - 1000}=\frac{18375 - 18000}{50}=\frac{375}{50}=7.5$.

Step2: Use the four - step process for differentiation

1. $R(x + h)=28(x + h)-0.010(x + h)^{2}=28x+28h-0.010(x^{2}+2xh + h^{2})=28x+28h-0.010x^{2}-0.020xh-0.010h^{2}$.
2. $R(x + h)-R(x)=(28x+28h-0.010x^{2}-0.020xh-0.010h^{2})-(28x - 0.010x^{2})=28h-0.020xh-0.010h^{2}$.
3. $\frac{R(x + h)-R(x)}{h}=\frac{28h-0.020xh-0.010h^{2}}{h}=28 - 0.020x-0.010h$.
4. $\lim_{h

ightarrow0}\frac{R(x + h)-R(x)}{h}=\lim_{h
ightarrow0}(28 - 0.020x-0.010h)=28 - 0.02x$. So, $R^{\prime}(x)=28 - 0.02x$.

Step3: Evaluate $R(x)$ and $R^{\prime}(x)$ at $x = 1000$

$R(1000)=28\times1000-0.010\times1000^{2}=18000$.
$R^{\prime}(1000)=28-0.02\times1000=28 - 20 = 8$.
The derivative $R^{\prime}(x)$ represents the instantaneous rate of change of the revenue function $R(x)$. A positive value of $R^{\prime}(1000)$ means the revenue is increasing.

Answer:

(A) 7.5
(B) $R^{\prime}(x)=28 - 0.02x$
(C) $R(1000)=18000$, $R^{\prime}(1000)=8$
B. This means that at a production level of 1,000 car seats, the revenue is $R(1000)$ dollars and is increasing at a rate of $R^{\prime}(1000)$ dollars per seat.