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svetlana [45]
3 years ago
7

Se desea construir un vaso de papel en forma de cono circular recto que tenga un volumen de 25πcm3

Mathematics
1 answer:
Genrish500 [490]3 years ago
5 0

Answer:

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

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

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 (A_{s}), en centímetros cuadrados, sea mínima para el volumen dado (V), en centímetros cúbicos. Las ecuaciones de volumen y área superficial son, respectivamente:

V = \pi\cdot r^{2}\cdot h (1)

A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r\cdot h (2)

De (1):

h = \frac{V}{\pi\cdot r^{2}}

En (2):

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

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

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

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

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

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

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

4\pi\cdot r - 2\cdot V\cdot r^{-2} = 0

4\pi\cdot r = 2\cdot V\cdot r^{-2}

4\pi\cdot r^{3} = 2\cdot V

r^{3} = \frac{V}{2\pi}

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

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

A_{s}'' = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }

A''_{s} = 4\pi + 8\pi

A_{s}'' = 12\pi (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):

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

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

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

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

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

r \approx 2.321\,cm

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

h \approx 4.642\,cm

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

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

r \approx 5.419\,cm

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

h \approx 10.839\,cm

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