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MATH2-Inf: 1b und 2b korrigiert.
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Jim Martens
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@ -26,7 +26,8 @@
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\makeatother
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\begin{document}
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\author{Jim Martens (6420323)}
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\author{Jan Branitz (6326955), Jim Martens (6420323),\\
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Stephan Niendorf (6242417)}
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\title{Hausaufgaben zum 21. Oktober}
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\maketitle
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\section{} %1
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@ -56,12 +57,28 @@
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\multicolumn{10}{r}{$x_{1}^{'}, x_{1}^{''}, x_{2}, x_{3}, x_{4}$} \,&\geq &\, 0
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\end{alignat*}
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\subsection{} %b
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Es gilt das folgende Problem mit der grafischen Methode zu lösen.
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\begin{alignat*}{3}
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\text{maximiere}\; & 2x_{1} &+& 5x_{2}&& \\
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\multicolumn{6}{l}{\text{unter den Nebenbedingungen}} && \\
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& 3x_{1} &-& 2x_{2} &\leq &\, 6 \\
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& x_{1} &+& x_{2} &\leq &\, 6 \\
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-& 2x_{1} &+& 6x_{2} &\leq &\, 18 \\
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\multicolumn{4}{r}{$x_{1}, x_{2}$} \,&\geq &\, 0
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\end{alignat*}
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Nach Umstellen der Nebenbedingungen nach $x_{2}$ ergibt sich dieses:
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\begin{alignat*}{3}
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x_{2} &\geq & \frac{3}{2}x_{1} &-& 3 \\
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x_{2} &\leq & -x_{1} &+& 6 \\
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x_{2} &\leq & \frac{1}{3}x_{1} &+& 3
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\end{alignat*}
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Daraus lässt sich die Fläche aller gültigen Werte zeichnen.
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\begin{tikzpicture}[>=stealth]
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\begin{axis}[
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ymin=0,ymax=7,
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x=1em,
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y=1em,
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x=1cm,
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y=1cm,
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axis x line=middle,
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axis y line=middle,
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axis line style=->,
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@ -72,13 +89,30 @@
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\addplot[no marks, black, -] expression[domain=2:6,samples=100]{1.5*x - 3} node[pos=0.65,anchor=north]{};
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\addplot[no marks, black, -] expression[domain=0:6,samples=100]{-1*x + 6} node[pos=0.65,anchor=north]{};
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\addplot[no marks, black, -] expression[domain=0:6,samples=100]{0.33333*x + 3} node[pos=0.65,anchor=north]{};
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\addplot[no marks, black, -] expression[domain=0:7,samples=100]{-0.4*x + 4.6} node[pos=0.65,anchor=north]{};
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%\node at (axis cs: 2.75,-3) {f};
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%\node at (axis cs: 1.75,-2.5) {A};
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\addplot[no marks, black, -] expression[domain=0:6,samples=100]{0.3333333333333*x + 3} node[pos=0.65,anchor=north]{};
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\addplot[no marks, black, -] expression[domain=0:7,samples=100]{-0.4*x + 4.65} node[pos=0.65,anchor=north]{};
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\node at (axis cs: 2.5,4.5) {(2.2,3.75)};
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\node at (axis cs: 6,2) {z};
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%\draw[>=stealth] (axis cs:1,0) -- (axis cs:1,-6) node [pos=0.65,anchor=north]{};
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\end{axis}
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\end{tikzpicture}
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\end{tikzpicture}\\
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Das optimale Ergebnis kann folgendermaßen bestimmt werden:
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\begin{alignat*}{5}
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I &-&\; 2x_{1} &+& 6x_{2} &=& 18 && \;| + 2II\\
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II &&\; x_{1} &+& x_{2} &=& 6 && \\
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\overset{I+2II}{\Rightarrow} &&\; && 8x_{2} &=& 30 && \;| \cdot \frac{1}{8} \\
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\Leftrightarrow &&\; && x_{2} &=& \frac{30}{8} = \frac{15}{4} &&
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\intertext{Einsetzen in II}
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\overset{II}{\Rightarrow} &&\; x_{1} &+& \frac{15}{4} &=& 6 &&\;| - \frac{15}{4} \\
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&&\; x_{1} && &=& \frac{24}{4} - \frac{15}{4} = \frac{9}{4} &&
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\end{alignat*}
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Anhand der beiden $x$-Werte kann nun der Wert der Zielfunktion berechnet werden.
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\[
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2 \cdot \frac{9}{4} + 5 \cdot \frac{15}{4} = \frac{18}{4} + \frac{75}{4} = \frac{93}{4} = 23,25
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\]
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Damit ist $\frac{93}{4}$ das optimale Ergebnis für die Zielfunktion $2x_{1} + 5x_{2}$ unter den gegebenen Nebenbedingungen.
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\section{} %2
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\subsection{} %a
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Pauls Diätproblem:\\
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@ -109,14 +143,14 @@
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\multicolumn{12}{r}{$x_{1}^{'}, x_{1}^{''}, x_{2}, x_{3}, x_{4}^{'}, x_{4}^{''}$} \,&\geq &\, 0
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\end{alignat*}
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\subsection{} %b
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\begin{alignat*}{10}
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\text{maximiere}\; & 5a_{1} &+& 13a_{2} &+& 8a_{3} &+& 9b_{1} &+& 15b_{2} &+& 12b_{3} &+& 5c_{1} &+& 14c_{2} &+& 10c_{3} && \\
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\multicolumn{20}{l}{\text{unter den Nebenbedingungen}} && \\
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& a_{1} &+& a_{2} &+& a_{3} && && && && && && &\leq &\, 400 \\
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& && && && b_{1} &+& b_{2} &+& b_{3} && && && &\leq &\, 480 \\
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& && && && && && && c_{1} &+& c_{2} &+& c_{3} &\leq &\, 230 \\
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& && a_{2} && && &+& b_{2} && && &+& c_{2} && &\leq &\, 420 \\
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& && && a_{3} && && &+& b_{3} && && &+& c_{3} &\leq &\, 250 \\
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\multicolumn{18}{r}{$a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}$} \,&\geq &\, 0
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\begin{alignat*}{7}
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\text{maximiere}\; & 13a_{1} &+& 8a_{2} &+& 15b_{1} &+& 12b_{2} &+& 14c_{1} &+& 10c_{2} && \\
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\multicolumn{14}{l}{\text{unter den Nebenbedingungen}} && \\
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& a_{1} &+& a_{2} && && && && &\leq &\, 400 \\
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& && && b_{1} &+& b_{2} && && &\leq &\, 480 \\
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& && && && && c_{1} &+& c_{2} &\leq &\, 230 \\
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& a_{1} && &+& b_{1} && &+& c_{1} && &\leq &\, 420 \\
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& && a_{2} && &+& b_{2} && &+& c_{2} &\leq &\, 250 \\
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\multicolumn{12}{r}{$a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}$} \,&\geq &\, 0
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\end{alignat*}
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\end{document}
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