Suppose that the price p, in thousand pesos, and the number of sales x(in hundreds) of a certain item can be modeled by the equation 5p+4x+px=100. Suppose also that the price is increasing at the rate of $200 per year. How fast is the quantity changing at the instant when the price of the item is $1500?

Question

Suppose that the price p, in thousand pesos, and the number of sales x(in hundreds) of a certain item can be modeled by the equation 5p+4x+px=100. Suppose also that the price is increasing at the rate of $200 per year. How fast is the quantity changing at the instant when the price of the item is $1500?

Answer 1

The rate of $200 per year is dp/dt,

so we need to differentiate with respect to t

5p+4x+px=100
5 dp/dt + 4 dx/dt + p dx/dt + x dp/dt = 0

we need x when p = 1.5
in the original:
5(1.5) + 4x + 1.5x = 100
5.5x = 92.5
x = 185/11

so we have: p = 1.5, x = 185/11 , dx/dt = .2, and dp/dt = ???

5 dp/dt + 4 dx/dt + p dx/dt + x dp/dt = 0
5dp/dt + 4(.2) + 1.5(.2) + (185/11)(dp/dt) = 0
factor out dp/dt, and finish the arithmetic.
remember that dp/dt will be in thousands of pesos per year.

Answer 2

To find how fast the quantity is changing at the instant when the price of the item is $1500, we need to differentiate the given equation with respect to time (t) and solve for dx/dt.

First, let's differentiate the equation with respect to time:

d/dt(5p + 4x + px) = d/dt(100)

Since the price is increasing at the rate of $200 per year, we know that dp/dt = 200. Therefore, we can rewrite the equation as:

5(dp/dt) + 4(dx/dt) + p(dx/dt) = 0

We are interested in finding dx/dt when the price (p) is equal to $1500. So, we substitute p = 1500 and dp/dt = 200 into the equation:

5(200) + 4(dx/dt) + (1500)(dx/dt) = 0

1000 + 4(dx/dt) + 1500(dx/dt) = 0

Now, simplify the equation:

2500(dx/dt) + 1000 = 0

Subtract 1000 from both sides:

2500(dx/dt) = -1000

Divide both sides by 2500:

dx/dt = -1000/2500

Simplify the fraction:

dx/dt = -0.4

Therefore, the quantity is changing at a rate of -0.4 (in hundreds) per year when the price of the item is $1500. This means that the quantity is decreasing.

Answer 3

To find how fast the quantity is changing at the instant when the price is $1500, we need to find the derivative of the quantity with respect to time.

First, let's rewrite the given equation:

5p + 4x + px = 100

Since p is the price in thousand pesos and x is the number of sales in hundreds, we can convert the equation to:

5000p + 400x + 1000px = 100000

Dividing by 1000 to simplify the equation, we get:

5p + 0.4x + px = 100

Now, let's find the derivative of both sides with respect to time:

d(5p)/dt + d(0.4x)/dt + d(px)/dt = d(100)/dt

Since the price is increasing at a rate of $200 per year, the derivative of p with respect to time (dp/dt) is 200.

So we have:

5(dp/dt) + 0.4(dx/dt) + p(dx/dt + x(dp/dt)) = 0

Now, let's substitute p = 1500 into the equation:

5(200) + 0.4(dx/dt) + 1500(dx/dt + x(200)) = 0

1000 + 0.4(dx/dt) + 1500(dx/dt + 200x) = 0

Next, let's solve for dx/dt:

0.4(dx/dt) + 1500(dx/dt) + 300000x = -1000

1900(dx/dt) + 300000x = -1000

Finally, we can substitute the given price p = 1500 to find the quantity x:

5(1500) + 4x + 1500x = 100

7500 + 4x + 1500x = 100

7500 + 1550x = 100

1550x = -7400

x = -4.77

Since the number of sales cannot be negative, we can ignore the negative value.

Therefore, at the instant when the price of the item is $1500, the quantity is not changing as it is negative.

Please note that it's important to double-check the calculations, as there may be a possibility of error.