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

Answer:

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

Step-by-step explanation:

Given function is, $8p^{3}+x^{2}=45225$

To determine the rate at which sales are changing, that is, $\dfrac{dx}{dt}$, differentiate given function with respect to t,  

$\dfrac{d}{dt}\left ( 8p^{3}+x^{2} \right )=\dfrac{d}{dt}\left ( 45225 \right )$

Applying sum rule of derivative,

$\dfrac{d}{dt}\left ( 8p^{3} \right )+\dfrac{d}{dt}\left ( x^{2} \right )=\dfrac{d}{dt}\left ( 45225 \right )$

Applying power rule and constant rule of derivative,

$8\left ( 3p^{3-1} \right )\dfrac{dp}{dt}+\left ( 2x^{2-1} \right )\dfrac{dx}{dt}=0$

$8\left ( 3p^{2} \right )\dfrac{dp}{dt}+\left ( 2x^{1} \right )\dfrac{dx}{dt}=0$

$8\left ( 3p^{2} \right )\dfrac{dp}{dt}+\left ( 2x \right )\dfrac{dx}{dt}=0$

$24\left ( p^{2} \right )\dfrac{dp}{dt}+2\left ( x \right )\dfrac{dx}{dt}=0$

Substituting the values,  $x=135,p=15,\dfrac{dp}{dt}= -\:0.60$

Since it is given that price is falling, so $\dfrac{dp}{dt}$ is negative.

$24\left ( 15^{2} \right )\left ( -\:0.60 \right )+2\left ( 135 \right )\dfrac{dx}{dt}=0$

$24\times225\times\left ( -\:0.60 \right )+270\:\dfrac{dx}{dt}=0$

$-3240+270\:\dfrac{dx}{dt}=0$

Adding -3240 from both sides,

$270\:\dfrac{dx}{dt}=3240$

Dividing by 270,

$\dfrac{dx}{dt}=\dfrac{3240}{270}$

$\dfrac{dx}{dt}=12$