Related Rates Questions! Need Help!!

Question

It is estimated that the annual advertising revenue received by a certain newspaper will be
R(x)=0.5x^2+4x+179
thousand dollars when its circulation is x thousand. The circulation of the paper is currently 11,000 and is increasing at a rate of 2,000 papers per year. At what rate will the annual advertising revenue be increasing with respect to time 3 years from now?

Advertising revenue will be increasing at a rate $     ?        thousand per year.

 

Suppose that the demand function q for a certain product is given by
q=9000e^(−0.005⋅p)
where p denotes the price of the product. If the item is currently selling for $200 per unit, and the quantity supplied is decreasing at a rate of 80 units per week, find the rate at which the price of the product is changing.

The price of the product is changing at       ?      dollars per week.

Answer 1

·We know that the rate of change of a function is the derivative of the function. Note that the function is given of x, but we are asked to find the rate of change with respect to time (t), knowing that x is a function of t. Therefore we need the chain rule to find the derivative.

R(x) = .5x²+4x+179

dR/dt = .5(2x)dx/dt +4dx/dt = (x+4)dx/dt

  We are told in the problem that dx/dt = 2000 papers per year. We also need to know the value of x in three years. If it is 11,000 now, and is increasing 2000 a year (which we must assume is a constant rate of change), then in 3 years it will be 11,000 + 6,000 = 17,000 papers. Be careful to note that x is measured in thousands, so we put 17 into the formula, not 17,000

 

So, dR/dt = (17 + 4)·2 = 42. Also note that R is given in thousands of dollars, so the rate of increase of the revenue is $42,000 per year at that point. Since the word "thousand" is given as part of the question, just 42 goes in place of the question mark.

 

For the second part, we are given that q = 9000e^-.005p.

We are also told that p is currently 200 and that dq/dt = -80 units/week.

We are asked to find dp/dt

Using implicit differentiation, we can find  dq/dt = d/dt(9000e^-.005p)

 

dq/dt = 9000(-.005pe^-.005pdp/dt)

 

We can plug in the values of dq/dt and p to get

 

-80=-45(200)e^-1dp/dt

 

-80 =(-9000/e)(dp/dt)

 

Dividing both sides by -9000/e, we get

 

80e/9000 = dp/dt

 

2e/225 = dp/dt

 

dp/dt is approximately equal to  $0.012  (or 1.2 cents) per week.