Find the rate of revenue change with respect to time.

Question

is

Mobile phone manufacture will be releasing a new version of their smart phone model. They

determined that t months after the new phone introduced to the market;

The manufacturing facility can produce x(t) hundred units.

The price of the unit p(t) will be changed by the time as well.

Given

t^2+3t hundreds of units produced where t is the month since the release time

p(t) = -2t^(3/2) + 30 dollars per unit.

The revenue R(t) = x(t) p(t)

At what rate is the revenue changing with respect to the time after 4 months?

Is the revenue increasing or decreasing at this time?

Answer 1

So, the problem wants you to determine the rate of revenue change R(t) with respect to time t when t=4 months.

I’d first determine the time derivative of the revenue function, i.e.

d / dt (R(t)) = d / dt (x(t)*p(t))

thinking of it this way you could apply the product rule of differentiation...

d / dt (R(t)) = x(t)* d / dt (p(t)) + p(t) * d / dt (x(t)).

then evaluate this result at t = 4.

so notice you’ll have to take the time derivative of both x(t) and p(t) to replace them in the equation above for d / dt (R(t)). Both will utilize the power rule of differentiation, i.e.

d / dx (xⁿ) = n*x^n-1.

For example, given p(t) = -2t^3/2+ 30, then

d / dt (p(t)) = -2*(3/2)t^1/2 + 0 = -3t^1/2.

I’ll leave d / dt (x(t)) for you to find.

Then substitute into the time derivative equation d / dt (R(t)) above. I’ll fill it out for everything but x(t) that I left for you.

d / dt (R(t)) = (t²+3t)(-3t^1/2) + (-2t^3/2+30) d / dt (x(t))

After substituting in your finding of d / dt (x(t)) to the above equation and simplifying, make the substitution t=4 and evaluate the RHS of the equation to determine the answer to the first question. The final question is answered by the sign of your findings to the first question, positive is increasing negative is decreasing.