Question

Calculus question!

When the price of a certain commodity is pp dollars per unit, consumers demand x hundred units of the commodity, where x^2+3px+p^2=79 How fast is the demand x changing with respect to time when the price is $5 per unit and is decreasing at the rate of 50 cents per month?

Answer 1

First things first, can you take the derivative of the equation?

Answer 2

\[\frac{ -2x-3p }{ 2p+3x }\]

Answer 3

@steve816

Answer 4

That is not quite right!

Answer 5

You must use implicit differentiation.

Answer 6

Implicit Differentiation:

2x * dx/dt + 3x * dp/dt + 3p * dx/dt + 2p * dp/dt = 0
--> solve for dx/dt (the quantity we want)

dx / dt * (2x + 3p) + 3x * dp/dt + 2p * dp/dt = 0
--> put everything without dx/dt on the right side

dx / dt * (2x + 3p) = -3x * dp/dt - 2p * dp/dt
--> divide by dx/dt's coefficient

dx/dt = -(3x * dp/dt + 2p * dp/dt) / (2x + 3p)
--> we are given p and dp/dt, now we need to find x using the original equation

Plug in p = 5 in the original equation to find x

x² + 3*5*x + 5² = 79
-->

x² + 15x + 25 = 79
--> set equal to zero...subtract 79 from both sides

x² + 15x - 54 = 0
--> first let's try to factor, find factors of 54 that subtract to give 15

2*27 = 54, 27 - 2 = 25, NOPE
3*18 = 54, 18 - 3 = 15, BINGO

(x + 18)(x - 3) = 0
--> set each factor to zero

x + 18 = 0 --> x = -18
x - 3 = 0 --> x = 3

Now, the demand probably CANNOT be negative, meaning the only reasonable solution is x = 3, now plug into our formula for dx/dt:

dx/dt = -(3*3 * (-0.30) + 2*5 * (-0.30)) / (2*3 + 3*5)

(notice that the units are dollars per unit, so decreasing @ 30¢ means a negative and 0.30 dollars)

dx/dt = -(9 * (-0.30) + 10 * (-0.30)) / (6 + 15)
-->

dx/dt = -(-2.7 + -3) / 21
-->

dx/dt = -(-5.7) / 21 = 5.7/21 ~ 0.271428571 hundred units per month

(so the demand is increasing...which sort of makes sense that the demand increases as the supply diminishes).


btw, you don't have to solve explicitly for dx/dt, you could wait and plug the numbers into the implicit differentiation equation once you know x, p, and dp/dt:

2x * dx/dt + 3x * dp/dt + 3p * dx/dt + 2p * dp/dt = 0
-->

2*3 * dx/dt + 3*3 * -.3 + 3*5 * dx/dt + 2*5 * -.3 = 0
--> if it makes things easier you could substitute X = dx/dt

6X - 2.7 + 15X - 3 = 0
--> collect like terms (the X's), add constants to both sides

21X = 5.7
-->

X = 5.7/21 ~ 0.271428571