to find the x value of the max of
f(x)=ax^2+bx+c
when a is negative (if a is positive, we find the minimum)
we do
-b/2a is the x value
to find the y value, we just sub that x value back into the function
so
R(x)=-0.2x^2+60x+0
-b/2a=-60/(2*0.2)=-60/-0.4=150
x value is 150
make 150 units
sub back to find revenue
R(150)=-0.2(150)^2+60(150)
R(150)=-0.2(22500)+9000
R(150)=-4500+9000
R(150)=4500
max revenue is achieved when 150 units are produced yeilding $4500 in revenue
