Question

Suppose that the price p​ (in dollars) and the weekly sales x​ (in thousands of​ units) of a certain commodity satisfy the demand equation 8p cubedplusx squaredequals45 comma 225. Determine the rate at which sales are changing at a time when xequals135​, pequals15​, and the price is falling at the rate of ​$.60 per week.

Answer 1

The value of rate at which the sales are changing is 12

Explanation:

Given function is,
`8p^(3)+x^(2)=45225`

To determine the rate at which sales are changing, that is,
`(dx)/(dt)`, differentiate given function with respect to t,

`(d)/(dt)\left ( 8p^(3)+x^(2) \right )=(d)/(dt)\left ( 45225 \right )`

Applying sum rule of derivative,

`(d)/(dt)\left ( 8p^(3) \right )+(d)/(dt)\left ( x^(2) \right )=(d)/(dt)\left ( 45225 \right )`

Applying power rule and constant rule of derivative,

`8\left ( 3p^(3-1) \right )(dp)/(dt)+\left ( 2x^(2-1) \right )(dx)/(dt)=0`

`8\left ( 3p^(2) \right )(dp)/(dt)+\left ( 2x^(1) \right )(dx)/(dt)=0`

`8\left ( 3p^(2) \right )(dp)/(dt)+\left ( 2x \right )(dx)/(dt)=0`

`24\left ( p^(2) \right )(dp)/(dt)+2\left ( x \right )(dx)/(dt)=0`

Substituting the values,
`x=135,p=15,(dp)/(dt)= -\:0.60`

Since it is given that price is falling, so
`(dp)/(dt)` is negative.

`24\left ( 15^(2) \right )\left ( -\:0.60 \right )+2\left ( 135 \right )(dx)/(dt)=0`

`24*225*\left ( -\:0.60 \right )+270\:(dx)/(dt)=0`

`-3240+270\:(dx)/(dt)=0`

Adding -3240 from both sides,

`270\:(dx)/(dt)=3240`

Dividing by 270,

`(dx)/(dt)=(3240)/(270)`

`(dx)/(dt)=12`