Answer:
No, there is no price for which the weekly revenue would be 1200
Step-by-step explanation:
If r = 1200
[tex]-4p^2+40p+887=1200\\\\\implies -4p^2+40p-313=0[/tex]
Therefore, [tex]a=-4, \ \ \ b=40, \ \ \ c=-313[/tex]
discriminant = [tex]b^2-4ac[/tex]
[tex]\implies 40^2-4(-4)(-313) = 1600 - 5008=-3408\\\\-3408<0 \implies \textsf{no real solutions}[/tex]