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	<id>https://wikiskola.se/index.php?action=history&amp;feed=atom&amp;title=L%C3%A5dan_-_en_praktisk_%C3%B6vning</id>
	<title>Lådan - en praktisk övning - Versionshistorik</title>
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	<updated>2026-05-11T15:53:57Z</updated>
	<subtitle>Versionshistorik för denna sida på wikin</subtitle>
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	<entry>
		<id>https://wikiskola.se/index.php?title=L%C3%A5dan_-_en_praktisk_%C3%B6vning&amp;diff=33481&amp;oldid=prev</id>
		<title>Hakan: /* Här är så en GGB som visar lösningen: */</title>
		<link rel="alternate" type="text/html" href="https://wikiskola.se/index.php?title=L%C3%A5dan_-_en_praktisk_%C3%B6vning&amp;diff=33481&amp;oldid=prev"/>
		<updated>2015-12-16T19:54:15Z</updated>

		<summary type="html">&lt;p&gt;&lt;span dir=&quot;auto&quot;&gt;&lt;span class=&quot;autocomment&quot;&gt;Här är så en GGB som visar lösningen:&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
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				&lt;td colspan=&quot;2&quot; style=&quot;background-color: #fff; color: #202122; text-align: center;&quot;&gt;← Äldre version&lt;/td&gt;
				&lt;td colspan=&quot;2&quot; style=&quot;background-color: #fff; color: #202122; text-align: center;&quot;&gt;Versionen från 16 december 2015 kl. 19.54&lt;/td&gt;
				&lt;/tr&gt;&lt;tr&gt;&lt;td colspan=&quot;2&quot; class=&quot;diff-lineno&quot; id=&quot;mw-diff-left-l73&quot;&gt;Rad 73:&lt;/td&gt;
&lt;td colspan=&quot;2&quot; class=&quot;diff-lineno&quot;&gt;Rad 73:&lt;/td&gt;&lt;/tr&gt;
&lt;tr&gt;&lt;td class=&quot;diff-marker&quot;&gt;&lt;/td&gt;&lt;td style=&quot;background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;&quot;&gt;&lt;div&gt;Du kan även testa var max hamnar om du flyttar glidarna till andra mått på papperet (a och b kan variera från noll till A4-storlek).&lt;/div&gt;&lt;/td&gt;&lt;td class=&quot;diff-marker&quot;&gt;&lt;/td&gt;&lt;td style=&quot;background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;&quot;&gt;&lt;div&gt;Du kan även testa var max hamnar om du flyttar glidarna till andra mått på papperet (a och b kan variera från noll till A4-storlek).&lt;/div&gt;&lt;/td&gt;&lt;/tr&gt;
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&lt;tr&gt;&lt;td colspan=&quot;2&quot; class=&quot;diff-side-deleted&quot;&gt;&lt;/td&gt;&lt;td class=&quot;diff-marker&quot; data-marker=&quot;+&quot;&gt;&lt;/td&gt;&lt;td style=&quot;color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;&quot;&gt;&lt;div&gt;&lt;ins style=&quot;font-weight: bold; text-decoration: none;&quot;&gt;&amp;lt;&lt;/ins&gt;/&lt;ins style=&quot;font-weight: bold; text-decoration: none;&quot;&gt;html&lt;/ins&gt;&amp;gt;&lt;/div&gt;&lt;/td&gt;&lt;/tr&gt;
&lt;/table&gt;</summary>
		<author><name>Hakan</name></author>
	</entry>
	<entry>
		<id>https://wikiskola.se/index.php?title=L%C3%A5dan_-_en_praktisk_%C3%B6vning&amp;diff=22183&amp;oldid=prev</id>
		<title>Hakan: Skapade sidan med &#039;== Uppgift att tillverka en låda med störst volym ==  Tänk dig att du formar en låda genom att klippa bort lika stora kvadrater i hörnen av ett rektangulärt papper och d...&#039;</title>
		<link rel="alternate" type="text/html" href="https://wikiskola.se/index.php?title=L%C3%A5dan_-_en_praktisk_%C3%B6vning&amp;diff=22183&amp;oldid=prev"/>
		<updated>2013-05-01T21:00:18Z</updated>

		<summary type="html">&lt;p&gt;Skapade sidan med &amp;#039;== Uppgift att tillverka en låda med störst volym ==  Tänk dig att du formar en låda genom att klippa bort lika stora kvadrater i hörnen av ett rektangulärt papper och d...&amp;#039;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Ny sida&lt;/b&gt;&lt;/p&gt;&lt;div&gt;== Uppgift att tillverka en låda med störst volym ==&lt;br /&gt;
&lt;br /&gt;
Tänk dig att du formar en låda genom att klippa bort lika stora kvadrater i hörnen av ett rektangulärt papper och därefter viker upp kanterna. Vad ska kvadraterna ha för sidlängd för att maximera lådans volym? Vi kan för klarhetens skull kalla papperssidorna a respektive b och kvadratens sida x.&lt;br /&gt;
&lt;br /&gt;
=== Vi bekantar oss med problemet. ===&lt;br /&gt;
&lt;br /&gt;
För att slippa detaljer och decimaler tänker vi oss ett papper som har sidorna 10 cm och 20 cm.&lt;br /&gt;
&lt;br /&gt;
# Om vi klipper bort 1 cm i varje hörn blir botten på lådan &amp;lt;math&amp;gt;8 * 18 cm^2&amp;lt;/math&amp;gt;. Volymen blir då &amp;lt;math&amp;gt;8 * 18 * 1 cm^3 = 144 cm^3&amp;lt;/math&amp;gt;.&lt;br /&gt;
#  Om vi klipper bort 2 cm i varje hörn blir botten på lådan &amp;lt;math&amp;gt;6 * 16 cm^2&amp;lt;/math&amp;gt;. Volymen blir då &amp;lt;math&amp;gt;6 * 16 * 2 cm^3 = 96 * 2 = 192 cm^3&amp;lt;/math&amp;gt;.&lt;br /&gt;
#  Om vi klipper bort 3 cm i varje hörn blir botten på lådan &amp;lt;math&amp;gt;4 * 14 cm^2&amp;lt;/math&amp;gt;. Volymen blir då &amp;lt;math&amp;gt;4 * 14 * 3 cm^3 = 56 * 3 = 168 cm^3&amp;lt;/math&amp;gt;.&lt;br /&gt;
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[[Fil:Ladan_test.png|300px|right]]&lt;br /&gt;
Nu ser vi att vi verkar ha en maxvolym om man klipper bort 2 cm. problemet är att vi inte vet om det är störst volym när man klipper bort exakt 2 cm. Det kan ju hända att lösningen är att klippa bort ett hörn med ett mått som anges i cm, mm och många decimaler därtill.&lt;br /&gt;
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Vi kan nu ta våra värden på x och volym och rita en enkel graf.&lt;br /&gt;
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Nu har vi begripit mer av hur problemet fungerar. Dags att vi ger oss in på det riktiga problemet.&lt;br /&gt;
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=== Lösning ===&lt;br /&gt;
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Papperets kanter har längderna a och b.&lt;br /&gt;
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Lådans kanter har längderna a-2x och b-2x efter det att man har klippt bort hörnen. Hörnen har ju sidan x.&lt;br /&gt;
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Man viker upp kanterna på lådan och kanterna är förstås x höga.&lt;br /&gt;
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Lådans volym ges då av bottenarean multiplicerat med kantens höjd. Vi kan skriva ett uttryck för lådans volyn V(x).&lt;br /&gt;
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: &amp;lt;math&amp;gt; V(x) = (a-2x)(b-2x) x &amp;lt;/math&amp;gt;&lt;br /&gt;
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För att få fram den maximala volyme ska vi derivera uttrycket för volymen och sätta derivatan lika med noll. Det kommer att ge oss en ekvation där lösningen är det x-värde som ger störstvolym.&lt;br /&gt;
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Vi börjar med att förenkla uttrycket genom att multiplicera ihop parenteserna, mm.&lt;br /&gt;
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: &amp;lt;math&amp;gt; V(x) = (a-2x)(b-2x) x  = abx - 2 ax^2 - 2bx^2 + x^3&amp;lt;/math&amp;gt;&lt;br /&gt;
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Om vi deriverar&lt;br /&gt;
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: &amp;lt;math&amp;gt; V(x) = abx - 2 ax^2 - 2bx^2 + x^3&amp;lt;/math&amp;gt;&lt;br /&gt;
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får vi &lt;br /&gt;
: &amp;lt;math&amp;gt; V&amp;#039;(x)  = ab - 4 ax - 4bx + 3x^2&amp;lt;/math&amp;gt;&lt;br /&gt;
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Då sätter vi derivatan lika med noll vilket betyder:&lt;br /&gt;
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: &amp;lt;math&amp;gt;  ab - 4 ax - 4bx + 3x^2 = 0&amp;lt;/math&amp;gt;  &lt;br /&gt;
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Det är en andragradsekvation som vi stuvar om så att den stämmer med pq-formeln:&lt;br /&gt;
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: &amp;lt;math&amp;gt;  x^2 - \frac {4}{3} (a - b) x + \frac {ab}{3}   = 0&amp;lt;/math&amp;gt;&lt;br /&gt;
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Lösningen är:&lt;br /&gt;
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: &amp;lt;math&amp;gt;  x =  \frac {2}{3} (a - b)  +- \sqrt{(\frac {2}{3} (a - b))^2 - \frac {ab}{3}}  &amp;lt;/math&amp;gt;&lt;br /&gt;
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Den roten som ger vårt maximum är. &lt;br /&gt;
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: &amp;lt;math&amp;gt;  x =  \frac {2}{3} (a - b)  - \sqrt{(\frac {2}{3} (a - b))^2 - \frac {ab}{3}}  &amp;lt;/math&amp;gt;&lt;br /&gt;
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Om man sätter a = 10 och b = 20 som i testet ovan får vi max vid &amp;lt;math&amp;gt; x = 5 - \frac {5}{\scrt{3}}  \cong 2.11 &amp;lt;/math&amp;gt;&lt;br /&gt;
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===== Tips: =====&lt;br /&gt;
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Wolfram Alpha ger en snygg lösning: med a = 20 och b = 10: http://www.wolframalpha.com/input/?i=maximum+%2820-2x%29%2810-2x%29x&lt;br /&gt;
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== Här är så en GGB som visar lösningen: ==&lt;br /&gt;
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I denna GGB kan du se hur max inträffar där derivatan är lika med noll samt att tangenten i den punkten är helt horisontell.&lt;br /&gt;
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Du kan även testa var max hamnar om du flyttar glidarna till andra mått på papperet (a och b kan variera från noll till A4-storlek).&lt;br /&gt;
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		<author><name>Hakan</name></author>
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