MIT_单变量微积分_10

野性酷女 2022-03-14 13:38 95阅读 0赞

### \`1. 最大值、最小值 ###

`Ex1:`  
把一个长度为1 的绳子分成两段，每段围成一个正方形，求所能得到的最大的正方形面积。  
 S = ( x 4 ) 2 + ( 1 − x 4 ) 2 S=(\\frac\{x\}\{4\})^2+(\\frac\{1-x\}\{4\})^2 S=(4x)2\+(41−x)2  
 S ′ = 0 S&\#x27;=0 S′=0  
 S ′ = x 8 − 1 − x 8 = 0 S&\#x27;=\\frac\{x\}\{8\}-\\frac\{1-x\}\{8\}=0 S′=8x−81−x=0  
 x = 1 2 x=\\frac\{1\}\{2\} x=21（驻点）.  
 S ( 1 2 ) = ( 1 8 ) 2 + ( 1 8 ) 2 = 1 32 S(\\frac\{1\}\{2\})=(\\frac\{1\}\{8\})^2+(\\frac\{1\}\{8\})^2=\\frac\{1\}\{32\} S(21)=(81)2\+(81)2=321  
 S ( 0 + ) = 0 + 1 16 = 1 16 S(0^+)=0+\\frac\{1\}\{16\}=\\frac\{1\}\{16\} S(0\+)=0\+161=161  
 S ( 1 − ) = 1 16 + 0 = 1 16 S(1^-)=\\frac\{1\}\{16\}+0=\\frac\{1\}\{16\} S(1−)=161\+0=161  
几何表示：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4Mzg2MzE2_size_16_color_FFFFFF_t_70]  
`Ex2:`  
一个无盖固定容积的盒子，使其表面积最小。  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4Mzg2MzE2_size_16_color_FFFFFF_t_70 1]  
**解：** v = x 2 y , y = v x 2 , A = x 2 + 4 x y = x 2 + 4 v x = 0. v=x^2y,y=\\frac\{v\}\{x^2\}, A=x^2+4xy=x^2+\\frac\{4v\}\{x\}=0. v=x2y,y=x2v,A=x2\+4xy=x2\+x4v=0.  
 x 3 = 2 v , x = 2 1 3 v 1 3 ( 驻 点 ) x^3=2v,x=2^\{\\frac\{1\}\{3\}\}v^\{\\frac\{1\}\{3\}\}(驻点) x3=2v,x=231v31(驻点)  
 ( 0 < x < ∞ ) (0<x<\\infty) (0<x<∞)  
 A ( 0 + ) = ( x 2 + 4 v x ) ∣ x = 0 + = ∞ A(0^\{+\})=(x^2+\\frac\{4v\}\{x\})|\_\{x=0^\{+\}\}=\\infty A(0\+)=(x2\+x4v)∣x=0\+=∞  
 A ( ∞ ) = ( x 2 + 4 v x ) ∣ x = ∞ = ∞ A(\\infty)=(x^2+\\frac\{4v\}\{x\})|\_\{x=\\infty\}=\\infty A(∞)=(x2\+x4v)∣x=∞=∞  
 A ′ ′ = 2 + 8 v x 3 > 0 A&\#x27;&\#x27;=2+\\frac\{8v\}\{x^3\}>0 A′′=2\+x38v>0  
 y = v x 2 = v 2 1 3 v 1 3 = v 1 3 2 1 3 y=\\frac\{v\}\{x^2\}=\\frac\{v\}\{2^\{\\frac\{1\}\{3\}\}v^\{\\frac\{1\}\{3\}\}\}=v^\{\\frac\{1\}\{3\}\}2^\{\\frac\{1\}\{3\}\} y=x2v=231v31v=v31231  
 A = x 2 + 4 x y = 2 7 2 v 2 3 A=x^2+4xy=2^\{\\frac\{7\}\{2\}\}v^\{\\frac\{2\}\{3\}\} A=x2\+4xy=227v32  
**结论**： x y = 2 \\frac\{x\}\{y\}=2 yx=2

*  使用隐函数求导法：  
     V = x 2 y , A = x 2 + 4 x y V=x^2y,A=x^2+4xy V=x2y,A=x2\+4xy  
     d d x ( v = x 2 y ) ⇒ 0 = 2 x y + x 2 y ′ \\frac\{d\}\{dx\}(v=x^2y) \\Rightarrow 0 = 2xy+x^2y&\#x27; dxd(v=x2y)⇒0=2xy\+x2y′  
     y ′ = − 2 x y x 2 = − 2 y x y&\#x27;=\\frac\{-2xy\}\{x^2\}=-\\frac\{2y\}\{x\} y′=x2−2xy=−x2y  
     d A d x = 2 x + 4 y + 4 x y ′ = 2 x − 4 y = 0 \\frac\{dA\}\{dx\}=2x+4y+4xy&\#x27;=2x-4y=0 dxdA=2x\+4y\+4xy′=2x−4y=0  
    得出： x y = 2 \\frac\{x\}\{y\}=2 yx=2

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4Mzg2MzE2_size_16_color_FFFFFF_t_70]: /images/20220313/5521d56a8c394cbc86d2ab7f1b9525f2.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4Mzg2MzE2_size_16_color_FFFFFF_t_70 1]: /images/20220313/cec728730a284b94801becba353692ce.png