Re: Beazley's Problem

ram@zedat.fu-berlin.de (Stefan Ram) writes:

  Alright, so here's how I approached it: We know that when the
  price x is 5 bucks, the number of people n is 120 (^1).

That assumption doesn't seem so good, but accepting it, your answer
looks right.  Here is a pure numerical solution.  Since the profit
function is quadratic, the Newton iteration converges immediately.
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def cost(n): return 180+.04*n   # cost to show to n viewers
def revenue(price,n): return price*n # amount collected from them
def people(price): return 120.+(price-5)*(-15./.1) # number who will attend
def profit(price):
    n = people(price)
    return revenue(price,n) - cost(n)

def ddx(f,x,h=0.001): return (f(x+h)-f(x-h))/(2*h) # numerical derivative
def newton(f,x0): return x0 - f(x0)/ddx(f,x0)      # Newton-Raphson iteration

def dprofit(price): return ddx(profit, price) # derivative of profit

x = 5.
for i in range(3):
    print(f'{i} {x:.4f} {profit(x):.1f} {dprofit(x):.1f}')
    x = newton(dprofit,x)