∂I/(∂h_2 )=W_2/√(〖(h_2^2+〖(20-x)〗^2)〗^3 )-3 (W_2 h_2^2)/√(〖(h_2^2+〖(20-x)〗^2)〗^5 ) =0
时间: 2024-03-07 14:52:30 浏览: 18
对 $I(x,h_2)$ 求关于 $h_2$ 的偏导数得:
$$\frac{\partial I}{\partial h_2}=\frac{W_2}{\sqrt{(h_2^2+(20-x)^2)^3}}-\frac{3W_2h_2^2}{\sqrt{(h_2^2+(20-x)^2)^5}}$$
要使 $\frac{\partial I}{\partial h_2}=0$,则需要解方程:
$$\frac{W_2}{\sqrt{(h_2^2+(20-x)^2)^3}}-\frac{3W_2h_2^2}{\sqrt{(h_2^2+(20-x)^2)^5}}=0$$
化简后得:
$$\frac{1}{h_2^2}=\frac{3}{h_2^2+(20-x)^2}$$
解得:
$$h_2= \pm \frac{20-x}{\sqrt{2}}$$
因为 $h_2$ 必须大于 $0$,所以:
$$h_2=\frac{20-x}{\sqrt{2}}$$
将其代入原式,可以求出最小值。
相关问题
用latex表示:■(∂C/∂x=3 (P_2 h_2 (s-x))/√(〖(h_2^2+(s-x)^2)〗^5 )-3 (P_1 h_1 x)/√(〖(h_1^2+x^2)5〗 )@∂C/(∂h_2 )=P_2/√(〖(h_2^2+(s-x)^2)〗^3 )-3 (P_1 h_2^2)/√(〖(h_2^2+(s-x)^2)〗^5 )) ∂C/(∂h_1 )=(P_2 h_1^2)/√(〖(h_1^2+x^2)〗^5 )+P_1/√(〖(h_1^2+x^2)〗^3 )
$$\frac{\partial C}{\partial x}=3\left(\frac{P_2h_2(s-x)}{\sqrt{(h_2^2+(s-x)^2)^5}}-\frac{P_1h_1x}{\sqrt{(h_1^2+x^2)^5}}\right)$$
$$\frac{\partial C}{\partial h_2}=\frac{P_2}{\sqrt{(h_2^2+(s-x)^2)^3}}-\frac{3P_1h_2^2}{\sqrt{(h_2^2+(s-x)^2)^5}}$$
$$\frac{\partial C}{\partial h_1}=\frac{P_2h_1^2}{\sqrt{(h_1^2+x^2)^5}}+\frac{P_1}{\sqrt{(h_1^2+x^2)^3}}$$
用latex表示:∂I/∂x=3 (W_2 h_2 (20-x))/√(〖(h_2^2+〖(20-x)〗^2)〗^5 )-3 (W_1 h_1 x)/√(〖(h_1^2+x^2)〗^5 )= (-30(20-√2 h))/√(〖(25+x^2)〗^5 )+(9h_2 (20-x))/√(〖(h_2^2+〖(20-x)〗^2)〗^5 )=0
$$\frac{\partial I}{\partial x} = 3\left(\frac{W_2h_2(20-x)}{\sqrt{\left(h_2^2+(20-x)^2\right)^5}} - \frac{W_1h_1x}{\sqrt{\left(h_1^2+x^2\right)^5}}\right) = -\frac{30(20-\sqrt{2}h)}{\sqrt{\left(25+x^2\right)^5}} + \frac{9h_2(20-x)}{\sqrt{\left(h_2^2+(20-x)^2\right)^5}} = 0$$
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