Raymond B. answered 11/11/22
Math, microeconomics or criminal justice
Profit = P = R-C
P= -.5x^2 +80x -4x-300
=-.5x^2 +76x -300
rewrite in vertex form
= -.5(x^2 -76x) -300
complete the square. add and subtract 76^2(.5) = 5776/2 = 2888
=-.5(x^2-138x+76^2) -300+2888
=-.5(x-76)^2 +2588
vertex is (76, 2588) = maximum point.
$2588 is max profit, 76 is the profit maximizing output level
answer to part c is yes, $2500 profit is feasible, but just barely
P=50 = -.5x^2 +76x -300
-.5x^2 +76x -300=50
-.5x^2 +76x=250
-.5(x^2 -132 + 76^2) = 250-76^2/2
-.5(x-76)^2 = -2638
(x-76)^2 = -2638/-.5 = 2638(2) = 5276
x-76 =about +/- 72.64
x = 3.36 or 148.64 are both output levels with close to $50 profit
the profit maximizing output level=76 is the midpoint between output levels with $50 profit
another method is use calculus
take the derivative of the profit function and set =0, solve for x
that's the same as setting Marginal Revenue = Marginal Cost
which gives the profit maximizing output level

Peter R.
11/11/22