Question 1032512
a) To break even, P = 0...we get
{{{-2x^2 + 14x - 20 = 0}}}
{{{x^2 - 7x + 10 = 0}}}
{{{(x-2)(x-5) = 0}}}
Thus they break even at 200 meals and at 500 meals.
b)  To maximize profit, we take the derivative and set it equal to zero, so that
P' = -4x + 14 = 0
and
x = 3.5 = 350 meals
c)  To find out what that profit is, we plug 3.5 in for x in the original and get
P(3.5) = -2(3.5)^2 + 14(3.5) - 20 = 4.5 = $4500