The demand for apples is given by the equation xp=200, where x is the number of pounds demanded and p is the price per pound. If the price is increasing at a rate of 1.25 dollars per week, when the...
The demand for apples is given by the equation xp=200, where x is the number of pounds demanded and p is the price per pound. If the price is increasing at a rate of 1.25 dollars per week, when the demand is 30 pounds, at what rate is the
a)demand changing=
b)revenue changing=
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
The price (p) and demand (d) for apples is related by x*p = 200.
x*p = 200
Take the derivative of both the sides.
`x*((dp)/dt) + p*(dx/dt) = 0`
=> `-(x/p)*((dp)/dt) = dx/dt`
If price is increasing at $1.25 per week, `(dp)/dt` = 1.25. When demand is 30 pounds, x = 30 and the price is 200/30 = 20/3.
The change in demand is `-(30/(20/3))*1.25` = `-(90/20)*1.25` = -5.625. The demand is decreasing at 5.625 pound/week.
Revenue is given by price*demand, R = x*p = $200. This is constant at 200.
At the given conditions, the decrease in demand is 5.625 pounds/week but revenue remains unchanged at $200.
pramodpandey | College Teacher | (Level 3) Valedictorian
Demand function
`p=200/x` (i)
differentiate with tespect to 't'
`(dp)/(dt)=-200/x^2 (dx)/(dt)`
given , `(dp)/(dt)=1.25 ` when x=30 ,therefore
`(dx)/(dt)=-(1.25xx30^2)/200`
`=-5.625`
Revenue function is
R(x)=px
=(200/x)x
R(x) =200 (ii)
differentiate (ii) with respect to x
R'(x)=0
Thus revenue will not change.