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A cardboard box manufacturing company is building boxes with length represented by x+ 1, width by 5- x, and height by x -1. The volume of the box is modeled by the function below V(x) 18 14 10 6 24 X 5 6 2 2 3 -2 -6 Over which interval is the volume of the box changing at the fastest average rate? [1,2] A. [1,3.5 B. C. [1,5] r0,3.51 D

User GoreDefex
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Answer:

a. [1,2]


m= (9-0)/(2-1)=9

b. [1,3.5]


m =(17-0)/(3.5-1)=6.8

c. [1,5]


m =(0-0)/(5-1)=0

d. [0,3.5]


m =(17-(-5))/(3.5-0)=6.29

So then we can conclude that the highest slope is for the interval [1,2] and that would be our solution for the fastest average rate.

a. [1,2]


m= (9-0)/(2-1)=9

Explanation:

Assuming that we have the figure attached for the function. For this case we just need to quantify the slope given by:


m = (\Delta y)/(\Delta x)

For each interval and the greatest slope would be the interval on which the volume of the box is changing at the fastest average rate

a. [1,2]


m= (9-0)/(2-1)=9

b. [1,3.5]


m =(17-0)/(3.5-1)=6.8

c. [1,5]


m =(0-0)/(5-1)=0

d. [0,3.5]


m =(17-(-5))/(3.5-0)=6.29

So then we can conclude that the highest slope is for the interval [1,2] and that would be our solution for the fastest average rate.

a. [1,2]


m= (9-0)/(2-1)=9

A cardboard box manufacturing company is building boxes with length represented by-example-1
User Nioq
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