我正在尝试将线性程序转换为标准形式,但我无法使用 & 将等号和不等号放在同一列中。如果我尝试添加更多 &,则会出现以下错误。
\documentclass[12pt]{article}
\begin{document}
\begin{eqnarray*}
&\min & x_1+3x_2-5x_3+2x_4-7x_5\\
&s.t. & 3x_2-8x_4+x_5=15\\
&&4x_1-8x_2+2x_4+x_5=25\\
&&x_1+8x_2-5x_4-29\geq 0\\
&&-2x_1-5x_2-5x_3+3x_5+27\geq 0\\
&&x_2, x_4 \geq 0\\
&&-x_5\geq 0
\end{eqnarray*}
\end{document}
答案1
我想你想要第一个解决方案,但我认为第二个解决方案看起来更好:
\documentclass[12pt]{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
\min {}&x_1+3x_2-5x_3+2x_4-7x_5\\
\textup{ s.t.} & \begin{aligned}[t] 3x_2-8x_4+x_5 & =15\\
4x_1-8x_2+2x_4+x_5 & =25\\
x_1+8x_2-5x_4-29 & \geq 0\\
-2x_1-5x_2-5x_3+3x_5+27 & \geq 0\\
x_2, x_4 & \geq 0\\
-x_5 & \geq 0
\end{aligned}
\end{align*}
\begin{align*}
\min {}&x_1+3x_2-5x_3+2x_4-7x_5\\[1ex] \textup{ s.t.} &\enspace \begin{array}[t] {|l} 3x_2-8x_4+x_5 =15\\
4x_1-8x_2+2x_4+x_5 =25\\
x_1+8x_2-5x_4-29 \geq 0\\
-2x_1-5x_2-5x_3+3x_5+27 \geq 0\\
x_2, x_4 \geq 0\\
-x_5 \geq 0
\end{array}
\end{align*}
\end{document}
答案2
我不确定我是否明白你想要什么:看看这是否可以做到。
该输出由以下代码生成:
% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly
% declare the paper format.
\usepackage[T1]{fontenc} % Not always necessary, but recommended.
% End of standard header. What follows pertains to the problem at hand.
\usepackage{amsmath}
\begin{document}
Text before the equation.
\begin{alignat*}{1}
\min\quad & x_1+3x_2-5x_3+2x_4-7x_5 \\
\text{s.t.}\quad &
\begin{aligned}[t]
3x_2-8x_4+x_5 &= 15\\
4x_1-8x_2+2x_4+x_5 &= 25\\
x_1+8x_2-5x_4-29 &\geq 0\\
-2x_1-5x_2-5x_3+3x_5+27 &\geq 0\\
x_2, x_4 &\geq 0\\
-x_5 &\geq 0
\end{aligned}
\end{alignat*}
Text after the equation.
\end{document}
还有一种可能性:
代码:
% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly
% declare the paper format.
\usepackage[T1]{fontenc} % Not always necessary, but recommended.
% End of standard header. What follows pertains to the problem at hand.
\usepackage{amsmath}
\begin{document}
Text before the equation.
\begin{alignat*}{1}
\min\quad & x_1+3x_2-5x_3+2x_4-7x_5 \\
\text{s.t.}\quad &
\left\{
\begin{aligned}
3x_2-8x_4+x_5 &= 15\\
4x_1-8x_2+2x_4+x_5 &= 25\\
x_1+8x_2-5x_4-29 &\geq 0\\
-2x_1-5x_2-5x_3+3x_5+27 &\geq 0\\
x_2, x_4 &\geq 0\\
-x_5 &\geq 0
\end{aligned}
\right.
\end{alignat*}
Text after the equation.
\end{document}
答案3
以下是使用[t]
op-aligned 的纯粹美学选项array
:
\documentclass{article}
\usepackage{lipsum}
\usepackage{amsmath}
\begin{document}
\lipsum*[1]
\begin{align*}
\min \quad & x_1 + 3 x_2 - 5 x_3 + 2 x_4 - 7 x_5 \\
\text{subject to} \quad &
\renewcommand{\arraystretch}{1.2}
\setlength{\arraycolsep}{0pt}
\begin{array}[t]{ *{7}{r} }
& {} 3 x_2 & & {} - 8 x_4 & {} + x_5 & {} = {} & 15 \\
4 x_1 & {} - 8 x_2 & & {} + 2 x_4 & {} + x_5 & {} = {} & 25 \\
x_1 & {} + 8 x_2 & & {} - 5 x_4 & & {}\geq {} & 29 \\
-2 x_1 & {} - 5 x_2 & {} - 5 x_3 & & {} + 3 x_5 & {}\geq {} & -27 \\
\multicolumn{5}{r}{x_2, x_4} & {}\geq {} & 0 \\
\multicolumn{5}{r}{x_5} & {}\leq {} & 0
\end{array}
\end{align*}
\lipsum[2]
\end{document}