CÁLCULO/4. DERIVADAS/5 Aplicaciones
RA7 4.4 Optimización: Minimizar material usado en cilindro con volumen dado.
e desea construir un envase con forma de cilindro circular recto. Este envase debe contener un volumen de #V1 .Si el fondo y la tapa tiene
doble espesor que la parte lateral del cilindro (como se observa en la figura, que representa la plantilla para construir dicho envase), entonces el valor del
radio que minimiza la cantidad de material es:
.
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Solución:
Para resolver el problema, notemos que la pregunta es hallar las dimensiones del radio que minimiza el cantidad de material. Para esto, empezamos definiendo las variables, sea «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»R«/mi»«/math» el radio de la base del cilindro y sea «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math» la altura del cilindro.
Con esto tenemos,
«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»
El problema pide minimizar material, y como se necesita doble espesor en la tapa y fondo del envase, tenemos que la función a minimizar es
«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»
Ahora, el volumen a contener es «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»#V1«/mi»«/math»; entonces,
«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»
Despejando la variable «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math» en la fórmula anterior:
«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»
Luego de despejar la variable «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math» , la reemplazamos en la ecuación de Área; quedando
«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»
Como el ejercicio pide minimizar material, debemos derivar la función área con respecto a «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»R«/mi»«/math»,
«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»
Igualamos a cero para obtener los puntos críticos:
«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»
Así, el punto crítico de la función área es «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»#R1«/mi»«/math».
Ahora, calculamos el signo que posee la derivada de la función área «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»A«/mi»«/math»
«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»
Por lo tanto, el valor del radio que minimiza la función costo total es «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»R«/mi»«mo»=«/mo»«mi»#R1«/mi»«/math». ]]>
1.0000000
1.0000000
0
true
true
abc
#sol
¡Excelente!]]>
#op1
Revisa el desarrollo para identificar tus errores. ¡Tú puedes!]]>
#op2
Revisa el desarrollo para identificar tus errores. ¡Tú puedes!]]>
#op3
Revisa el desarrollo para identificar tus errores. ¡Tú puedes!]]>
<question><wirisCasSession><session lang="es" version="2.0"><library closed="false"><mtext style="color:#ffc800" xml:lang="es">variables</mtext><group><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V00</mi><mo>=</mo><mi>aleatorio</mi><mfenced><mrow><mn>1</mn><mo>,</mo><mn>10</mn></mrow></mfenced></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V0</mi><mo>=</mo><msup><mi>V00</mi><mn>3</mn></msup></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V1</mi><mo>=</mo><mi>sustituir_cadena</mi><mfenced><mrow><mo>&quot;</mo><mo>#</mo><mn>1</mn><mfenced><mrow><mo>#</mo><mn>2</mn></mrow></mfenced><mo>&quot;</mo><mo>,</mo><mi>V0</mi><mo>,</mo><mfenced><msup><mi>metro</mi><mn>3</mn></msup></mfenced></mrow></mfenced></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m2</mi><mo>=</mo><mi>sustituir_cadena</mi><mfenced><mrow><mo>&quot;</mo><mfenced><mrow><mo>#</mo><mn>1</mn></mrow></mfenced><mo>&quot;</mo><mo>,</mo><mfenced><msup><mi>metro</mi><mn>2</mn></msup></mfenced></mrow></mfenced></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V2</mi><mo>=</mo><mn>2</mn><mo>*</mo><mi>V0</mi></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V3</mi><mo>=</mo><mn>3</mn><mo>*</mo><mi>V0</mi></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R0</mi><mo>=</mo><mfrac><mrow><mn>2</mn><mo>*</mo><mi>V0</mi></mrow><mrow><mn>8</mn><mo>*</mo><mi>π</mi></mrow></mfrac></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R1</mi><mo>=</mo><mi>sustituir_cadena</mi><mfenced><mrow><mo>&quot;</mo><mroot><mrow><mo>#</mo><mn>1</mn></mrow><mn>3</mn></mroot><mfenced><mrow><mo>#</mo><mn>2</mn></mrow></mfenced><mo>&quot;</mo><mo>,</mo><mi>R0</mi><mo>,</mo><mfenced><mi>metro</mi></mfenced></mrow></mfenced></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"/></input></command><maction actiontype="comment"><comment><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>alternativas</mi></math></comment></maction><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>op01</mi><mo>=</mo><mn>12</mn><mo>*</mo><mi>V0</mi><mo>*</mo><mroot><msup><mi>R0</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mn>3</mn></mroot></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>op02</mi><mo>=</mo><mfrac><mrow><mn>3</mn><mo>*</mo><mi>V0</mi></mrow><mrow><mn>4</mn><mi>π</mi></mrow></mfrac><mo>*</mo><mroot><msup><mi>R0</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mn>3</mn></mroot></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>op03</mi><mo>=</mo><mfrac><mn>125</mn><mrow><mi>π</mi><mo>*</mo><mroot><mi>R0</mi><mn>3</mn></mroot></mrow></mfrac></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>sol</mi><mo>=</mo><mi>R1</mi></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>op1</mi><mo>=</mo><mi>sustituir_cadena</mi><mfenced><mrow><mo>&quot;</mo><mo>#</mo><mn>1</mn><mfenced><mrow><mo>#</mo><mn>2</mn></mrow></mfenced><mo>&quot;</mo><mo>,</mo><mi>op01</mi><mo>,</mo><mfenced><mi>metro</mi></mfenced></mrow></mfenced></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>op2</mi><mo>=</mo><mi>sustituir_cadena</mi><mfenced><mrow><mo>&quot;</mo><mo>#</mo><mn>1</mn><mfenced><mrow><mo>#</mo><mn>2</mn></mrow></mfenced><mo>&quot;</mo><mo>,</mo><mi>op02</mi><mo>,</mo><mfenced><mi>metro</mi></mfenced></mrow></mfenced></math></input></command><command><input><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>op3</mi><mo>=</mo><mi>sustituir_cadena</mi><mfenced><mrow><mo>&quot;</mo><mo>#</mo><mn>1</mn><mfenced><mrow><mo>#</mo><mn>2</mn></mrow></mfenced><mo>&quot;</mo><mo>,</mo><mi>op03</mi><mo>,</mo><mfenced><mi>metro</mi></mfenced></mrow></mfenced></math></input></command></group></library></session></wirisCasSession><localData><data name="cas">false</data></localData></question>