Question

A model for a company's revenue from selling a software package is R=-2.5p^2+500p, where p is the price in dollars of the software. What price will maximize revenue? Find the maximum revenue.

Answer 1

it is quadratic equation

Answer 2

u can use derivative to find max also

Answer 3

find dR/dp

Answer 4

and put it =0

Answer 5

\[R=\frac{ -5p^{2} }{ 2 }+500p\] \[\frac{ dR }{ dp }=-5p+500\] \[\frac{ dR }{ dp }=5(-p+100)\]
Let dR/dp=0\[5(p-100)=0\]
\[p=100\]
\[\frac{ d^{2}R}{ dp^{2} }=-5=f''(x)\]\[f''(100)=-5<0\]
Thus f''(100) is less than zero, thus f(100) is maximum by second derivative test.
\[f(100)=\frac{ -5(100^{2}) }{ 2 }+500(100)\] \[f(100)=5*10^{4}(\frac{ -1 }{ 2 }+1)\]\[f(100)=\frac{ 5*10^{4} }{ 2 }=2.5*10^{4}=25*10^3=25,000\]