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Use differentials to approximate the change in revenue corresponding to an increase in sales (or production) of one unit. Then compare this with the actual change in revenue. Function x-Value R = 48x − 1.5x2 x = 15

dR = ___ dollars
ΔR = ____ dollars

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Final answer:

To approximate the change in revenue when sales increase by 1 unit, we can use differentials. By finding the derivative of the revenue function and evaluating it at the initial sales level, we can determine the change in revenue. In this case, the revenue decreases by $27 for every increase of 1 unit in sales. The actual change in revenue can be calculated by subtracting the revenue at the increased sales level from the revenue at the initial sales level, resulting in a change of $1.5.

Step-by-step explanation:

To approximate the change in revenue corresponding to an increase in sales, we can use differentials. The function given is R = 48x - 1.5x^2, where x represents the number of units sold. We need to find dR, which is the differential of R with respect to x. To do this, we can find the derivative of R with respect to x and evaluate it at x = 15. The derivative of R with respect to x is dR/dx = 48 - 3x. Evaluating this at x = 15, we get dR = 48 - 3(15) = -27. This means that for every increase of 1 unit in sales, the revenue decreases by $27.

The actual change in revenue, denoted as ΔR, can be found by subtracting the revenue at the initial sales level from the revenue at the increased sales level. We need to find R for x = 16 and x = 15. Substituting x = 16 into the function R = 48x - 1.5x^2, we get R = 48(16) - 1.5(16)^2 = 768 - 384 = $384. Substituting x = 15, we get R = 48(15) - 1.5(15)^2 = 720 - 337.5 = $382.5. Therefore, ΔR = $384 - $382.5 = $1.5.

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