Se desea construir un vaso de papel en forma de cono circular recto que tenga un volumen de 25πcm3
. Determine las dimensiones del vaso que requiere menos papel. Cuales deben ser si el volumen
es de 1 litro?

Answers

Answer 1

Answer:

Para un vaso de [tex]V = 25\pi\,cm^{3}[/tex], las dimensiones del vaso son [tex]r \approx 2.321\,cm[/tex] y [tex]h \approx 4.642\,cm[/tex].

Para un vaso de [tex]V = 1000\,cm^{3}[/tex], las dimensiones del vaso son [tex]r \approx 5.419\,cm[/tex] y [tex]h \approx 10.839\,cm[/tex].

Step-by-step explanation:

El vaso se puede modelar como un cilindro recto. El enunciado pregunta por las dimensiones del vaso tal que su área superficial ([tex]A_{s}[/tex]), en centímetros cuadrados, sea mínima para el volumen dado ([tex]V[/tex]), en centímetros cúbicos. Las ecuaciones de volumen y área superficial son, respectivamente:

[tex]V = \pi\cdot r^{2}\cdot h[/tex] (1)

[tex]A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r\cdot h[/tex] (2)

De (1):

[tex]h = \frac{V}{\pi\cdot r^{2}}[/tex]

En (2):

[tex]A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot \left(\frac{V}{\pi\cdot r} \right)[/tex]

[tex]A_{s} = 2\cdot \left(\pi\cdot r^{2}+V\cdot r^{-1} \right)[/tex]

Asumamos que [tex]V[/tex] es constante, la primera y segunda derivadas de la función son, respectivamente:

[tex]A'_{s} = 2\cdot (2\pi\cdot r -V\cdot r^{-2})[/tex]

[tex]A'_{s} = 4\pi\cdot r - 2\cdot V\cdot r^{-2}[/tex] (3)

[tex]A''_{s} = 4\pi + 4\cdot V \cdot r^{-3}[/tex] (4)

Si igualamos [tex]A'_{s}[/tex] a cero, entonces hallamos los siguientes puntos críticos:

[tex]4\pi\cdot r - 2\cdot V\cdot r^{-2} = 0[/tex]

[tex]4\pi\cdot r = 2\cdot V\cdot r^{-2}[/tex]

[tex]4\pi\cdot r^{3} = 2\cdot V[/tex]

[tex]r^{3} = \frac{V}{2\pi}[/tex]

[tex]r = \sqrt[3]{\frac{V}{2\pi} }[/tex] (5)

Ahora, si aplicamos este valor a (4), tenemos que:

[tex]A_{s}'' = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }[/tex]

[tex]A''_{s} = 4\pi + 8\pi[/tex]

[tex]A_{s}'' = 12\pi[/tex] (6)

De acuerdo con este resultado, el valor crítico está asociado al área superficial mínima. Ahora, la altura se calcula a partir de (5) y (1):

[tex]h = \frac{V}{\pi\cdot \left(\frac{V}{2\pi} \right)^{2/3} }[/tex]

[tex]h = \frac{2^{2/3}\cdot \pi^{2/3}\cdot V}{\pi\cdot V^{2/3}}[/tex]

[tex]h = \frac{2^{2/3}\cdot V^{1/3}}{\pi^{1/3}}[/tex]

Si [tex]V = 25\pi\,cm^{3}[/tex], entonces las dimensiones del vaso son:

[tex]r = \sqrt[3]{\frac{25\pi\,cm^{3}}{2\pi} }[/tex]

[tex]r \approx 2.321\,cm[/tex]

[tex]h = \frac{2^{2/3}\cdot (25\pi\,cm^{3})^{1/3}}{\pi^{1/3}}[/tex]

[tex]h \approx 4.642\,cm[/tex]

Un litro equivale a 1000 centímetros cúbicos, las dimensiones del vaso son:

[tex]r = \sqrt[3]{\frac{1000\,cm^{3}}{2\pi} }[/tex]

[tex]r \approx 5.419\,cm[/tex]

[tex]h = \frac{2^{2/3}\cdot (1000\,cm^{3})^{1/3}}{\pi^{1/3}}[/tex]

[tex]h \approx 10.839\,cm[/tex]


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Answers

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trust me bro

The value of the tangent of ∠Q is 0.96.

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One area of mathematics known as trigonometry examines the relationship between the sides and angles of a right triangle. The relationship between sides and angles is defined for 6 trigonometric functions.

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Answer:

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Step-by-step explanation:

Given

The attached rhombus

Required

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Cauculate the shaded area.

Sorry I could not get the image

Answers

Answer:

Area of shaded part = 113 mm² (Approx.)

Step-by-step explanation:

Given:

Diameter of Semi-circle = 24 mm

Find:

Area of shaded part

Computation:

Radius of semi-cicrle = Diameter of Semi-circle / 2

Radius of semi-cicrle = 24 / 2

Radius of semi-cicrle = 12 mm

So,

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Area of shaded part = 113.04

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