Question

In Karla’s Discount Supermarket, the weekly demand for Mellow Yellow Tea satisfies the equation px + 12p =
600, where x is the weekly demand in pounds and p is the price in dollars per pound. Right now the price is
$6 per pound, but is decreasing at the rate of 8 cents per week. At what rate is the demand changing? At
what rate is the weekly revenue changing?

Answer 1

This sounds like a calculus problem.

they give you dp/dt (change in price with respect to time) as 0.08 dollars
they want to find dx/dt (change in demand with respect to time)

take the derivative with respect to time of
px +12p = 600 (remember that both p and x are variables. use the product rule for p*x)

once you have the derivative, replace p with 6, dp/dt with 0.08, and x (solve for x using the original equation with p=6)
now solve for dx/dt

Answer 2

ill try that right now! thanks!

Answer 3

ok so taking the derivative i got

p + x = -12
6 + x = -12
x = -18

im a little confused about the dp/dt with 0.08 part

Answer 4

remember:
\[ \frac{d}{dt} x= \frac{dx}{dt} \]
For example:
\[ \frac{d}{dt} xp= x \frac{d}{dt}p+p \frac{d}{dt}x \]

Answer 5

is this correct
6 dp/dt (0.08)(-18)+ dx/dt = 0

Answer 6

how did you get that?

can you first do
\[\frac{d}{dt} (x p +12 p =600)\]

Answer 7

p + x +12 = 0

Answer 8

no

let's do the derivative of 12 p WITH RESPECT TO P:
\[ \frac{d}{dp}12 p = 12\frac{dp}{dp} = 12\]
That is what you appeared to do.
But you want the derivative WI

Answer 9

but you want the derivative with respect to t:
\[ \frac{d}{dt} 12p = 12\frac{dp}{dt} \]

Answer 10

can you try again?

Answer 11

man im sorry i have to leave for class i was writing something but i think it was wrong also i appreciate the help tho

Answer 12

between my posts you almost have the entire derivative:
\[\frac{d}{dt} xp= x \frac{d}{dt}p+p \frac{d}{dt}x\]
and
\[\frac{d}{dt} 12p = 12\frac{dp}{dt}\]
put those together to get the derivative of
px + 12p =600

I should fix dp/dt to be negative, because p is decreasing
\[ \frac{dp}{dt} = -0.08\]