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Consider the following problem:
The total revenue Addison expects from selling rose bouquets in June changes at a rate of 
r(x)=8-0.5 x thousands of dollars (where 
x is the price per bouquet in dollars). When 
x=30, the total expected revenue is 
$2000 dollars. By how much does the total expected revenue change between a selling price of 
$30 and 
$40 ?
Which expression can we use to solve the problem?
Choose 1 answer:
(A) 
int_(30)^(40)r^(')(x)dx
(B) 
2000+int_(30)^(40)r(x)dx
(C) 
2000+int_(30)^(40)r^(')(x)dx
(D) 
int_(30)^(40)r(x)dx

Consider the following problem:\newlineThe total revenue Addison expects from selling rose bouquets in June changes at a rate of r(x)=80.5x r(x)=8-0.5 x thousands of dollars (where x x is the price per bouquet in dollars). When x=30 x=30 , the total expected revenue is $2000 \$ 2000 dollars. By how much does the total expected revenue change between a selling price of $30 \$ 30 and $40 \$ 40 ?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) 3040r(x)dx \int_{30}^{40} r^{\prime}(x) d x \newline(B) 2000+3040r(x)dx 2000+\int_{30}^{40} r(x) d x \newline(C) 2000+3040r(x)dx 2000+\int_{30}^{40} r^{\prime}(x) d x \newline(D) 3040r(x)dx \int_{30}^{40} r(x) d x

Full solution

Q. Consider the following problem:\newlineThe total revenue Addison expects from selling rose bouquets in June changes at a rate of r(x)=80.5x r(x)=8-0.5 x thousands of dollars (where x x is the price per bouquet in dollars). When x=30 x=30 , the total expected revenue is $2000 \$ 2000 dollars. By how much does the total expected revenue change between a selling price of $30 \$ 30 and $40 \$ 40 ?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) 3040r(x)dx \int_{30}^{40} r^{\prime}(x) d x \newline(B) 2000+3040r(x)dx 2000+\int_{30}^{40} r(x) d x \newline(C) 2000+3040r(x)dx 2000+\int_{30}^{40} r^{\prime}(x) d x \newline(D) 3040r(x)dx \int_{30}^{40} r(x) d x
  1. Integrate rate of change: To find the change in total expected revenue between a selling price of $30\$30 and $40\$40, we need to integrate the rate of change of revenue, r(x)r(x), from x=30x=30 to x=40x=40. This will give us the total change in revenue over that interval.
  2. Calculate revenue function: The rate of change of revenue is given by r(x)=80.5xr(x)=8-0.5x. We need to integrate this function from x=30x=30 to x=40x=40 to find the total change in revenue.
  3. Use correct expression: The correct expression to use for this problem is the integral of r(x)r(x) from 3030 to 4040, which is represented by 3040r(x)dx\int_{30}^{40} r(x) \, dx. This corresponds to option (D).
  4. Calculate integral: Now we calculate the integral of r(x)r(x) from 3030 to 4040.3040(80.5x)dx\int_{30}^{40} (8 - 0.5x) dx=[8x0.25x2]3040= [8x - 0.25x^2]_{30}^{40}=(8400.25402)(8300.25302)= (8\cdot40 - 0.25\cdot40^2) - (8\cdot30 - 0.25\cdot30^2)=(320400)(240225)= (320 - 400) - (240 - 225)=80(15)= -80 - (-15)=80+15= -80 + 15=65= -65
  5. Find total revenue change: The total change in expected revenue between a selling price of $30\$30 and $40\$40 is $65-\$65 thousand dollars (since the revenue is given in thousands of dollars).

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