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第二单元用python学习微积分（十一）最值问题下和相关变率

2022-01-09 18:57:59 作者：互联网

本文内容来自于学习麻省理工学院公开课：单变量微积分-相关变率-网易公开课

一、最值问题举例

1、将一根长度为1的线，切成2段，每一段圈成一个正方形，求所能得到的最大面积

$f(x) =\frac{x^2}{16} + \frac{(1-x)^2}{16} = \frac{x^2 +1-2x+x^2}{16}| = \frac{x^2 -x+\frac{1}{2}}{8}_{x<1}$

$f'(x) =(\frac{x^2 -x+\frac{1}{2}}{8})' = \frac{1}{4}x - \frac{1}{8}$

$f'(x) = \frac{1}{4}x - \frac{1}{8}|_{x=\frac{1}{2}} =0$ , $f(\frac{1}{2}) = \frac{1}{32}$

计算两端：

$f(x)|_{x=1} = \frac{1}{16}, f(x)|_{x=0} = \frac{1}{16}$

$f'(x) = \frac{1}{4}x - \frac{1}{8}|_{x>\frac{1}{2}} >0$ , 驻点所以满足条件时应该x越大函数取值越大，当x->1时，函数最大 $f(x)|_{x\rightarrow1} = \frac{1}{16}$ $f(x)|_{x\rightarrow1} = \frac{1}{16}$

$f'(x) = \frac{1}{4}x - \frac{1}{8}|_{x<\frac{1}{2}} <0$ , 驻点所以满足条件时应该x越小函数取值越大，当x->0时，函数最大 $f(x)|_{x\rightarrow0} = \frac{1}{16}$

考虑两边高中间低这种驻点，所以驻点处得到最小面积值 1/32

from sympy import *
import numpy as np 

import matplotlib.pyplot as plt 

fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
ax.spines['left'].set_position('zero')
ax.spines['bottom'].set_position('zero')
ax.spines['right'].set_color('none')
ax.spines['top'].set_color('none')
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')
ax.set_aspect(10 ) 

def DrawXY(xFrom,xTo,steps,expr,color,label,plt):
    yarr = []
    xarr = np.linspace(xFrom ,xTo, steps) 
    for xval in xarr:
        yval = expr.subs(x,xval)
        yarr.append(yval)
    y_nparr = np.array(yarr) 
    plt.plot(xarr, y_nparr, c=color, label=label)    

def TangentLine(exprY,x0Val,xVal):
    diffExpr = diff(exprY)
    x1,y1,xo,yo = symbols('x1 y1 xo yo')
    expr = (y1-yo)/(x1-xo) - diffExpr.subs(x,x0Val)
    eq = expr.subs(xo,x0Val).subs(x1,xVal).subs(yo,exprY.subs(x,x0Val))
    eq1 = Eq(eq,0)
    solveY = solve(eq1)
    return xVal,solveY

def DrawTangentLine(exprY, x0Val,xVal1, xVal2, clr, txt):
    x1,y1 = TangentLine(exprY, x0Val, xVal1)
    x2,y2 = TangentLine(exprY, x0Val, xVal2)
    if len(txt)>0:
        plt.plot([x1,x2],[y1,y2], color = clr, label=txt)
    else:
        plt.plot([x1,x2],[y1,y2], color = clr)
        
x= symbols('x')
y = x*x/16 + (1-x)*(1-x)/16
DrawXY(0,1,100,y,'green','x*x/16 + (1-x)*(1-x)/16',plt)
plt.plot([0,1],[1/16,1/16], color = 'gray', label= 'y=1/16')

plt.legend(loc='lower right')
plt.show()

2、找到固定容积的无顶盖的盒子，使其表面积的最小值, (提示：底部是正方形--- 由于有正方形周长最短)

$V= x^2 y$

此处y可以称约束, $y = \frac{V}{x^2}$

$A = x^2+ 4xy$ (没有顶)

$A = x^2 + 4x\frac{V}{x^2} = x^2 +4\frac{V}{x}$

A' = 0

$A' = 2x - \frac{4V}{x^2} .......$ $2x - \frac{4V}{x^2} = 0......$ $x^3 = 2V ......$ $x = 2^{\frac{1}{3}} V^{\frac{1}{3}}$ (驻点）

端点 $0<x<\infty$

$A(0^+) = \infty$

$A(\infty) = \infty$

因为驻点为 $x = 2^{\frac{1}{3}} V^{\frac{1}{3}}$ ，所以图应该是

考虑二阶导数

$A'' = 4+\frac{8V}{x^3}$ (由x>0, 式子恒正，因此图形应该是上凹，而驻点应该是最小点）

最优解为： $y = \frac{V}{(2^{\frac{1}{3}}V^{\frac{1}{3}})^2} = 2^{-\frac{2}{3}}V^{\frac{1}{3}} ;$ $A = x^2 +4\frac{V}{x} = (2^{\frac{1}{3}} V^{\frac{1}{3}})^2 + 4(\frac{V}{2^{\frac{1}{3}} V^{\frac{1}{3}}}) =2^{\frac{2}{3}}V^{\frac{2}{3}} + 2\times2^{\frac{2}{3}}V^{\frac{2}{3}} = 3\times2^{\frac{2}{3}}V^{\frac{2}{3}}$ (我算的答案和视频中不同....)