Question 824418
{{{R(p)=-5p^2+1330p}}}
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{{{-5p^2+1330p=0}}} will have a maximum.
For this maximum,
{{{p=(-1330+- sqrt(1330^2-4*(-5)*0))/(2*(-5))}}}
{{{p=(-1330+- sqrt(1330))/(-10)}}}
{{{p=133+- (1/10)sqrt(1330)}}}
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Looking at the middle of the two possible p values,
{{{p=133/2}}} is where the maximum revenue should be.


{{{highlight(R(133/2)=-5*(133/2)^2+1330(133/2))}}}, maximum revenue.