Asked by Alyssa
                The demand equation for a product is 
p=88,650/(396+3x)
where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 <= x <= 50. (Round your answer to two decimal places.)
Please show work because I know the integral goes from 40 to 50 but I'm lost with the differentiation portion... Thank you
            
        p=88,650/(396+3x)
where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 <= x <= 50. (Round your answer to two decimal places.)
Please show work because I know the integral goes from 40 to 50 but I'm lost with the differentiation portion... Thank you
Answers
                    Answered by
            Damon
            
    integral p dx
of 88650 dx (3x+396)^-1
let z = 3x+396
then dz = 3 dx
and I have
88650 (dz/3)/z
or
29550 dz/z
= 29550 ln z
if x = 40, z = 120+396 = 516
if x = 50, z = 150+396 = 546
so
29550 [ln 546 - ln 516]
= 29550 (.0565) = 1670
divide by 50-40 = 10
p average = 167
=====================
CHECK
what is p(45)?
p(45) = 88650/(396+3(45))
= 167
LOL, close enough !
    
of 88650 dx (3x+396)^-1
let z = 3x+396
then dz = 3 dx
and I have
88650 (dz/3)/z
or
29550 dz/z
= 29550 ln z
if x = 40, z = 120+396 = 516
if x = 50, z = 150+396 = 546
so
29550 [ln 546 - ln 516]
= 29550 (.0565) = 1670
divide by 50-40 = 10
p average = 167
=====================
CHECK
what is p(45)?
p(45) = 88650/(396+3(45))
= 167
LOL, close enough !
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