Question 64540
{{{R = - (p^2) + 50p - 125}}}
use the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
where {{{R = a*p^2 + b*p + c}}}
a = -1
b = 50
c = -125
{{{p = (-50 +- sqrt( 50^2-4*(-1)*(-125)))/(2*(-1)) }}}
{{{p = (-50 +- sqrt( 2500 - 500)/ -2) }}}
{{{p = (-50 +- sqrt(2000)) / -2) }}}
{{{p = (-50 +- 20*sqrt(5)) / -2) }}}
{{{p = (25 +- 10*sqrt(5)) }}}
{{{p = 25 + 22.361}}}
{{{p = 47.361}}}
{{{p = 25 - 22.361}}}
{{{p = 2.639}}}
This says that when Brian charges $2.64, he has no income and when
he charges $47.36 he haas no income. This just says that if he charges too much, nobody wants to pay it. If he charges too little, he can't even make up for
his expenses. If he charges right in between, his income is maximum.
Because of the -1 coefficient of p^2, the parabola has a peak at he 
vertex where p = 25. Find R when p = 25
{{{R = - (25^2) + 50*25 - 125}}}
{{{R = - 625 + 1250 - 125}}}
{{{R = 500}}}
The maximum is at (25,500) He should charge $25 to make the max income
of $500
sketch the graph
{{{ graph( 550, 300, -10, 100, -30, 600, -(x^2) + 50*x - 125) }}}
How can Brian earn $400 /wk?
{{{400 = - (p^2) + 50p - 125}}}
From the graph, it looks like p = 14 would give about R = 400
{{{R = -(14^2) + 50*14 - 125}}}
{{{R = -196 + 700 -125}}}
{{{R = 379}}}
I'll try 14.5
{{{R = -(14.5^2) + 50*14.5 - 125}}}
{{{R = -210.25 + 725 -125}}}
{{{R = 389.75}}}
my next try would be p = 15 and that gives me R = 400
I'm not sure what the last part is.