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IgorC [24]
2 years ago
9

Suponga que desea utilizar un servicio de correo particular para enviar un paquete que tiene forma de caja rectangular con una s

ección transversal cuadrada tal que la suma de su longitud y el perímetro de la sección transversal es 100 pulgada, el máximo permitido por el servicio. (a) Encuentre un modelo matemático que exprese el volumen de la caja como una función de su longitud. (b) ¿Cuál es el dominio de la función del inciso (a)? (c) Determine en la graficadora, con aproximación de pulgadas, las dimensiones del paquete que tiene el mayor volumen posible que pueda enviarse por este servicio.
Mathematics
1 answer:
statuscvo [17]2 years ago
6 0

a) El volumen de la caja en función de su longitud es:

V_{caja}=\frac{x^{3}}{16}-\frac{25x^{2}}{2}+625x

b) El dominio de la ecuación del volumen son todos los números reales.

c) Las dimensiones del paquete con el mayor volumen posible son:

longitud de la caja  x = 30 plg

lado de la sección transversal L = 17.5 plg

a)

Definamos x como la longitud de la caja y L como el lado de la sección transversal, que es cuadrada para este caso.

Sabemos que la suma de su longitud (x) y el perímetro de la sección transversal (P = 4L) es igual a 100 plg.

x+4L=100 (1)

Ahora, el volumen de esta caja rectangular está dada por:

V_{caja}=A_{base}*x

V_{caja}=L^{2}*x (2)

Pero necesitamos expresar el volumen en función de x.

Despejamos L de la ecuación (1) y remplazarlo en (2).

Por lo tanto el volumen en función de x será.

V_{caja}=(\frac{100-x}{4})^{2}*x

V_{caja}=\frac{x^{3}}{16}-\frac{25x^{2}}{2}+625x

b)

Al ser la función un polinomio de orden 3 el dominio de esta función son todos los numeros reales.

c)

Observando la gráfica de esta función, podemos ver que para un valor aproxiamdo de x = 30 plg el valor de V tiene un punto de inflección, es por definición de maximización de una función que podemos usar ese punto para encontrar las dimensiones de la caja.

Por lo tanto si x = 30 plg el valor de L usando la ecuacion (1) sera:

L=\frac{100-x}{4}

L=\frac{100-30}{4}=17.5\: plg

Por lo tanto la máxima dimensión de la caja es:

x = 30 plg

L = 17.5 plg

Puedes aprender más sobre maximizar funciones aquí:

brainly.com/question/16339052

 

     

 

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A truck weighing 150kN and travelling at 2m/sec impacts with a buffer
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The maximum compression of the spring is 0.276 m

Step-by-step explanation:

First of all, we calculate the spring  constant of the spring, by using Hooke's law:

F=kx

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F = 10 kN = 10,000 N is the force applied

x = 1.25 cm = 0.0125 m is the corresponding compression of the spring

k is the spring constant

Solving for k,

k=\frac{F}{x}=\frac{10,000}{0.0125}=8\cdot 10^5 N/m

When the truck impacts with the spring, all the kinetic energy of the truck is converted into elastic potential energy of the spring, so we can write

\frac{1}{2}mv^2=\frac{1}{2}kx^2 (1)

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m is the mass of the truck

v = 2 m/s is the initial speed of the truck

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The mass of the truck can be found from its weight:

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So now we can re-arrange eq.(1) to find the compression, x:

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