1. ## Revenue problem

Revenue for sales of rubber baby-buggy bumpers is given by
R(q)=90q^2-q^3 for 0<q<70 .

a.what is the maximum revenue?
b.at what quantity is revenue increasing the fastest?

I tried finding R'(q)=180q-3q^2
and then from here found out p=-b/2a so that price(p) is set in R'(p) to find the maximum revenue.

I am not quite sure what part b wants.Am I doing part a correct?Any help?

2. Originally Posted by mathcalculushelp
Revenue for sales of rubber baby-buggy bumpers is given by
R(q)=90q^2-q^3 for 0<q<70 .

a.what is the maximum revenue?
b.at what quantity is revenue increasing the fastest?

I tried finding R'(q)=180q-3q^2
and then from here found out p=-b/2a so that price(p) is set in R'(p) to find the maximum revenue.

180q - 3q^2 = 0

3q(60 - q) = 0

q = 0 , q = 60 ... which value of q maximizes R ?

once you determine that fact, find R(q) for that value.

I am not quite sure what part b wants.Am I doing part a correct?Any help?

the question is asking for the value of q for which R'(q) is a maximum ...

R''(q) = 180 - 6q = 0

q = 30 ... will this value of q yield a maximum for R'(q) ?

how can you tell?
...

3. I'm not sure where you got $p=-\frac{b}{2a}$, but you are correct that we start by finding the first derivative of $R$. As $R$ is differentiable, its maximum will occur either at a boundary point or a point at which

$R'=180q-3q^2=0.$

Disregarding the solution $q=0$ and dividing, we obtain

$180-3q=0$

for the nonzero solution.

For the second problem, we are asked to find the rate of maximum increase of $R$, which amounts to finding the maximum of $R'$ over the interval $[0,70]$. As $R'$ is differentiable, we are again left with boundary points and points at which

$\frac{d}{dq}R'=\frac{d}{dq}(180q-3q^2)=180-6q=0.$