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S<i><font size="4" face="times new roman,times,serif">e desea construir <font size="4">un envase con forma de cil<font size="4">i</font>ndro <font size="4">circ<font size="4">ular recto. <font size="4">Este envase debe contener un volumen de <span class="nolink">«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»#V1«/mi»«/math»</span>. Si el fondo y la tapa tiene dob<font size="4">le espesor que la parte lat<font size="4">eral del cilindro, entonces el valor del radio que minimiza la cantidad de material es:</font></font></font></font></font></font></font></i><br /> <p align="justify"><br /><br /></p>
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<i><font size="4" face="times new roman,times,serif">Solución:<br /><br />Para resolver el problema, notemos que la pregunta es hallar las dimensiones del radio que minimiza el cantidad de material<font size="4">. </font>Para esto, empezamos definiendo las variables, sea <span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»R«/mi»«/math»</span> el radio de la base del cilindro y sea <span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»h«/mi»«/math»</span> la altura del cilindro. <br />Con esto tenemos,<br /><span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mtable»«mtr»«mtd»«mi»Á«/mi»«mi»r«/mi»«mi»e«/mi»«mi»a«/mi»«mo»§nbsp;«/mo»«mi»c«/mi»«mi»i«/mi»«mi»l«/mi»«mi»í«/mi»«mi»n«/mi»«mi»d«/mi»«mi»r«/mi»«mi»i«/mi»«mi»c«/mi»«mi»a«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mi»§#960;«/mi»«mi»R«/mi»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»Á«/mi»«mi»r«/mi»«mi»e«/mi»«mi»a«/mi»«mo»§nbsp;«/mo»«mi»f«/mi»«mi»o«/mi»«mi»n«/mi»«mi»d«/mi»«mi»o«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mi»Á«/mi»«mi»r«/mi»«mi»e«/mi»«mi»a«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»a«/mi»«mi»p«/mi»«mi»a«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«/mtable»«/math»</span></span><br />El problema pide minimi<font size="4">zar material<font size="4">, y c</font></font>omo <font size="4">se necesita dob<font size="4">le espesor en la tapa y fondo del envase, <font size="4">tenemos que la función a minimizar es </font></font></font><br /><span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mtable»«mtr»«mtd»«mi»Á«/mi»«mi»r«/mi»«mi»e«/mi»«mi»a«/mi»«/mtd»«mtd»«mo»:«/mo»«/mtd»«mtd»«mi»A«/mi»«mo»=«/mo»«/mtd»«/mtr»«/mtable»«mn»2«/mn»«mi»§#960;«/mi»«mi»R«/mi»«mi»h«/mi»«mo»+«/mo»«mn»4«/mn»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/math»</span></span><br />Ahora, el volumen a contener es <span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#V1«/mi»«/math»</span></span>; entonces, <br /><span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#V0«/mi»«mo»=«/mo»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«mi»h«/mi»«/math»</span></span><br />Despejando la variable <span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»h«/mi»«/math»</span> en la fórmula anterior: <br /><span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mtable»«mtr»«mtd»«mi»#V0«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»#V0«/mi»«/mrow»«mrow»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»h«/mi»«/mtd»«/mtr»«/mtable»«/math»</span></span><br />Luego de despejar la variable <span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»h«/mi»«/math»</span> , la reemplazamos en la ecuación de<font size="4"> Área</font>; quedando<br /><span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mtable»«mtr»«mtd»«mi»C«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mi»§#960;«/mi»«mi»R«/mi»«mo»·«/mo»«mfenced»«mfrac»«mi»#V0«/mi»«mrow»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfrac»«/mfenced»«mo»+«/mo»«mn»4«/mn»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mi»C«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»#V2«/mi»«mi»R«/mi»«/mfrac»«mo»+«/mo»«mn»4«/mn»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mi»C«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»#V2«/mi»«mi»R«/mi»«/mfrac»«mo»+«/mo»«mn»4«/mn»«mi»§#960;«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«/mtable»«/math»</span></span><br />Como el ejercicio pide minimizar material, debemos derivar la función área con respecto a <span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»R«/mi»«/math»</span>,<br /><span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mtable»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»R«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mfenced»«mrow»«mfrac»«mrow»«mi»#V2«/mi»«/mrow»«mi»R«/mi»«/mfrac»«mo»+«/mo»«mn»4«/mn»«mi»§#960;«/mi»«mi»R«/mi»«/mrow»«/mfenced»«/mrow»«mrow»«mi»d«/mi»«mi»R«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»R«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»-«/mo»«mfrac»«mi»#V2«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mfrac»«mo»+«/mo»«mn»8«/mn»«mi»§#960;«/mi»«mi»R«/mi»«/mtd»«/mtr»«/mtable»«/math»</span></span><br />Igualamos a cero para obtener los puntos críticos:<br /><span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mtable»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»R«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd»«mo»-«/mo»«mfrac»«mi»#V2«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mfrac»«mo»+«/mo»«mn»8«/mn»«mi»§#960;«/mi»«mi»R«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd»«mn»8«/mn»«mi»§#960;«/mi»«mi»R«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»#V2«/mi»«msup»«mi»R«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«msup»«mi»R«/mi»«mn»3«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»#V3«/mi»«mrow»«mn»8«/mn»«mi»§#960;«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»R«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mroot»«mi»#R0«/mi»«mn»3«/mn»«/mroot»«/mtd»«/mtr»«/mtable»«/math»</span></span><br />Así, el punto crítico de la función área es <span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»#R1«/mi»«/math»</span></span>.<br />Ahora, calculamos el signo que posee la derivada de la función <font size="4">área </font><span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»A«/mi»«/math»</span></span><br /><span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mtable»«mtr»«mtd»«mi»S«/mi»«mi»i«/mi»«mo»§nbsp;«/mo»«mn»0«/mn»«mo»§lt;«/mo»«mi»x«/mi»«mo»§lt;«/mo»«mroot»«mi»#R0«/mi»«mn»3«/mn»«/mroot»«/mtd»«mtd»«mo»§#8658;«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»R«/mi»«/mrow»«/mfrac»«mo»§lt;«/mo»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»S«/mi»«mi»i«/mi»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»§#10878;«/mo»«mroot»«mi»#R0«/mi»«mn»3«/mn»«/mroot»«/mtd»«mtd»«mo»§#8658;«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»R«/mi»«/mrow»«/mfrac»«mo»§gt;«/mo»«mn»0«/mn»«/mtd»«/mtr»«/mtable»«/math»</span></span><br />Por lo tanto, el valor del radio que minimiza la función costo total es <span class="nolink"><span class="nolink">«math xmlns="http://www.w3.org/1998/Math/MathML"»«mi»R«/mi»«mo»=«/mo»«mi»#R1«/mi»«/math»</span></span>. </font></i>
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<font size="4" face="times new roman,times,serif"><em>¡Excelente!</em></font>
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#op1
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<p align="justify"><font size="4" face="times new roman,times,serif"><em>Revisa el desarrollo para identificar tus errores. ¡Tú puedes!</em></font></p>
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<em><font size="4" face="Times New Roman">Revisa el desarrollo para identificar tus errores. ¡Tú puedes!</font></em>
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#op3
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<em><font size="4" face="Times New Roman">Revisa el desarrollo para identificar tus errores. ¡Tú puedes!</font></em>
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