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Re: Beazley's Problem

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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.
================================================================
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)

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Re: Beazley's Problem Paul Rubin <no.email@nospam.invalid> - 2024-09-21 05:45 -0700
  Re: Beazley's Problem Paul Rubin <no.email@nospam.invalid> - 2024-09-21 13:19 -0700
    Re: Beazley's Problem Annada Behera <annada@tilde.green> - 2024-09-23 13:14 +0530
      Re: Beazley's Problem (Posting On Python-List Prohibited) Lawrence D'Oliveiro <ldo@nz.invalid> - 2024-09-23 22:44 +0000
      Re: Beazley's Problem Paul Rubin <no.email@nospam.invalid> - 2024-09-23 17:22 -0700
        Re: Beazley's Problem Annada Behera <annada@tilde.green> - 2024-09-24 13:55 +0530
        Re: Beazley's Problem dkcombs@panix.com (david k. combs) - 2024-11-10 20:48 +0000
          Re: Beazley's Problem Paul Rubin <no.email@nospam.invalid> - 2024-11-10 13:55 -0800
      Re: Modern Optimization (was: Beazley's Problem) Gilmeh Serda <gilmeh.serda@nothing.here.invalid> - 2024-09-26 16:13 +0000
      Re: Beazley's Problem Antoon Pardon <antoon.pardon@vub.be> - 2024-10-06 22:19 +0200

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