Bruce Y. answered • 10/16/15

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·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) = .5x

^{2}+4x+179dR/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

^{-.005p}dp/dt)We can plug in the values of dq/dt and p to get

-80=-45(200)e

^{-1}dp/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.