Question

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

Answer 1

Answer:

a. [1,2]

$m= \frac{9-0}{2-1}=9$

b. [1,3.5]

$m =\frac{17-0}{3.5-1}=6.8$

c. [1,5]

$m =\frac{0-0}{5-1}=0$

d. [0,3.5]

$m =\frac{17-(-5)}{3.5-0}=6.29$

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]

$m= \frac{9-0}{2-1}=9$

Step-by-step explanation:

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

$m = \frac{\Delta y}{\Delta x}$

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]

$m= \frac{9-0}{2-1}=9$

b. [1,3.5]

$m =\frac{17-0}{3.5-1}=6.8$

c. [1,5]

$m =\frac{0-0}{5-1}=0$

d. [0,3.5]

$m =\frac{17-(-5)}{3.5-0}=6.29$

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]

$m= \frac{9-0}{2-1}=9$