Wiki-Quellcode von Lösung Fluß
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author | version | line-number | content |
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13.1 | 1 | [[image:Fluss.PNG||width="280" style="float: right"]] |
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7.1 | 2 | |
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10.1 | 3 | __Gegeben:__ {{formula}} \overline{AD}= 500\text{m}; \overline{BC}= 1000\text{m};{{/formula}} |
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12.1 | 4 | Geschwindigkeit von {{formula}}A{{/formula}} nach {{formula}}D{{/formula}}: {{formula}}v_{AD}= 50 \frac{\text{m}}{\text{min}}{{/formula}}; |
5 | Geschwindigkeit von {{formula}}D{{/formula}} nach {{formula}}C{{/formula}}: {{formula}}v_{DC}= 300 \frac{\text{m}}{\text{min}{{/formula}} | ||
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7.1 | 6 | |
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13.1 | 7 | __Gesucht:__ {{formula}}x{{/formula}} |
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7.1 | 8 | |
9 | Da der Sportler den Weg von {{formula}}D{{/formula}} zu {{formula}}C{{/formula}} 6 mal so schnell zurücklegt, wie den von {{formula}}A{{/formula}} zu {{formula}}D{{/formula}}, lautet die Hauptbedingung: | ||
10 | {{formula}}S = 6 \cdot \overline{AD} + \overline {DC}{{/formula}} | ||
11 | |||
12 | Die Nebenbedingungen lauten: | ||
13 | {{formula}}\overline{AD}= \sqrt{500^2+x^2}{{/formula}} | ||
14 | {{formula}}\overline{DC}= 1000 - x{{/formula}} | ||
15 | |||
16 | Somit lautet die Zielfunktion: | ||
17 | {{formula}}S(x)= 6 \cdot \sqrt{500^2+x^2} + 1000 - x {{/formula}} | ||
18 | |||
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13.1 | 19 | mit den Ableitungen |
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7.1 | 20 | |
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13.1 | 21 | {{formula}}S'(x)= \frac{6x}{\sqrt{500^2+x^2}}-1{{/formula}} |
22 | {{formula}}S''(x)= 6x \bigl(-\frac{1}{2}(500^2+x^2)^{-\frac{3}{2}}\cdot 2x \bigl) + 6(500^2+x^2)^{-\frac{1}{2}}{{/formula}} | ||
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7.1 | 23 | |
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13.1 | 24 | Durch die notwendige Bedingung {{formula}}S'(x)=0{{/formula}} ergibt sich |
25 | {{formula}} | ||
26 | \begin{align*} | ||
27 | \frac{6x}{\sqrt{500^2+x^2}}-1=0 \mid +1\\ | ||
28 | \frac{6x}{\sqrt{500^2+x^2}}= 1 \mid \cdot \sqrt{500^2+x^2}\\ | ||
29 | 6x = \sqrt{500^2+x^2} \mid ()^2 \\ | ||
30 | 36x^2= 500^2+x^2 \mid -x^2 \\ | ||
31 | 35x^2 = 500^2 \mid :35 \\ | ||
32 | x^2 = \frac{500^2}{35} \mid \sqrt \\ | ||
33 | x_1,2 = \pm \frac{100\sqrt{35}}{7} | ||
34 | \end{align*} | ||
35 | {{/formula}} |