# 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=

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

Posted on

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.