In [1]:

    

from math import sqrt


    

In [2]:

    

R = 100
A = 20
B = 30


    

In [3]:

    

def f(x):
    return sqrt(A**2 + x**2) + sqrt(B**2 + (R-x)**2)


    

In [13]:

    

# https://en.wikipedia.org/wiki/Golden-section_search
gr = (sqrt(5) + 1) / 2

def gss(f, a, b, c, tau = 1e-6):
    '''
    Python recursive version of Golden Section Search algorithm.

    This code appears to be broken - see the talk page.

    tau is the tolerance for the minimal value of function f
    b is any number between the interval a and c
    '''
    if b < c:
        x = b + (2 - gr) * (c - b)
    else:
        x = b - (2 - gr) * (b - a)
    if abs(c - a) < tau * (abs(b) + abs(x)): 
        return (c + a) / 2
    if f(x) < f(b):
        return gss(f, b, x, c, tau) if b < c else gss(f, a, x, b, tau)
    else:
        return gss(f, a, b, x, tau) if b < c else gss(f, x, b, c, tau)


    

In [5]:

    

gss(f, 0, R/2, R)


    

    
Out[5]:





40.0146627487199

In [6]:

    

import matplotlib.pyplot as plt


    

In [7]:

    

x = list(range(1,100))
plt.plot(x, [f(c) for c in x])
plt.show()


    

In [8]:

    

import sympy
from sympy import init_printing, simplify, factor, expand, cse, srepr, cancel
from sympy.abc import a,b,p,r
from sympy.physics.vector import vlatex
init_printing(latex_printer=vlatex, latex_mode='equation')


    

In [9]:

    

total_pipe = sympy.sqrt(a**2 + p**2) + sympy.sqrt(b**2 + (r-p)**2)
total_pipe


    

    
Out[9]:





$$\sqrt{a^{2} + p^{2}} + \sqrt{b^{2} + \left(- p + r\right)^{2}}$$

In [10]:

    

total_pipe_diff = sympy.diff(total_pipe, p)
total_pipe_diff


    

    
Out[10]:





$$\frac{p}{\sqrt{a^{2} + p^{2}}} + \frac{p - r}{\sqrt{b^{2} + \left(- p + r\right)^{2}}}$$

In [11]:

    

total_pipe_diff = total_pipe_diff.subs([(a, A),(b, B),(r, R)])
total_pipe_diff


    

    
Out[11]:





$$\frac{p}{\sqrt{p^{2} + 400}} + \frac{p - 100}{\sqrt{\left(- p + 100\right)^{2} + 900}}$$

In [12]:

    

total_pipe_diff_solve = sympy.solve(total_pipe_diff, p)
total_pipe_diff_solve


    

    
Out[12]:





$$\left [ 40\right ]$$