revenue function It is determined that $q$ units of a commodity can be sold when the price is $p$ hundred dollars per unit, where: $$ q(p)= 1000\cdot(p+2)e^{-p} $$ **Questions:** > a) For what price $p$ is revenue ...--prophetes.ai

revenue function It is determined that $q$ units of a commodity can be sold when the price is $p$ hundred dollars per unit, where: $$ q(p)= 1000\cdot(p+2)e^{-p} $$ Questions: > a) For what price $p$ is revenue $R$ maximized? > > b) What is the maximum revenue?

The revenue function is

$$ R(p) = p\cdot q = 1000p(p+2)\exp(-p) $$

The revenue is maximized when $R^\prime (p)=0$, that is,

$$ -1000(p^2 - 2)\exp(-p)=0 $$

Since $\exp(-p)$ will never vanish completely, the revenue is maximized when $p=\sqrt{2}$, as the price cannot be negative.

You may also want to conduct a second derivative test to ensure that it is indeed the (local) maximum.

revenue function It is determined that $q$ units of a commodity can be sold when the price is $p$ hundred dollars per unit, where: $$ q(p)= 1000\cdot(p+2)e^{-p} $$ **Questions:** > a) For what price $p$ is revenue $R$ maximized? > > b) What is the maximum revenue?

revenue function It is determined that $q$ units of a commodity can be sold when the price is $p$ hundred dollars per unit, where: $$ q(p)= 1000\cdot(p+2)e^{-p} $$ Questions: > a) For what price $p$ is revenue $R$ maximized? > > b) What is the maximum revenue?