Consider the following problem:The total revenue Addison expects from selling rose bouquets in June changes at a rate of r(x)=8−0.5x 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) ∫3040r(x)dx(B) 2000+∫3040r(x)dx(C) 2000+∫3040r′(x)dx(D) ∫3040r′(x)dx
Q. Consider the following problem:The total revenue Addison expects from selling rose bouquets in June changes at a rate of r(x)=8−0.5x 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) ∫3040r(x)dx(B) 2000+∫3040r(x)dx(C) 2000+∫3040r′(x)dx(D) ∫3040r′(x)dx
Understand the problem: Understand the problem.We need to find the change in total expected revenue as the price per bouquet increases from $30 to $40. The rate of change of revenue with respect to the price is given by the function r(x)=8−0.5x. To find the total change in revenue, we need to integrate this rate of change over the interval from x=30 to x=40.
Identify the correct expression: Identify the correct expression to use.To find the total change in revenue, we integrate the rate of change function r(x) over the interval from x=30 to x=40. The correct expression to use is the definite integral of r(x) from 30 to 40, which is represented by ∫3040r(x)dx. This corresponds to option (A).
Evaluate the integral to find the change in revenue: Evaluate the integral to find the change in revenue.We calculate the integral of r(x) from 30 to 40.∫3040r(x)dx=∫3040(8−0.5x)dx=[8x−0.25x2]3040=(8(40)−0.25(40)2)−(8(30)−0.25(30)2)=(320−400)−(240−225)=−80−(−15)=−80+15=−65The result is in thousands of dollars, so the change in revenue is 300.
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