2 a cardboard box manufacturing company is building boxes with length represented by x + 1, width by 5 - x…

2 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. over which interval is the volume of the box changing at the fastest average rate? 1) 1,2 2) 1,3.5 3) 1,5 4) 0,3.5 (1,0) (2,6) x1 y1 x2 y2 m = 6 - 0 / 2 - 1 = 6 / 1 = 6

2 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. over which interval is the volume of the box changing at the fastest average rate? 1) 1,2 2) 1,3.5 3) 1,5 4) 0,3.5 (1,0) (2,6) x1 y1 x2 y2 m = 6 - 0 / 2 - 1 = 6 / 1 = 6

Answer

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = V(x)$ over the interval $[a,b]$ is given by $\frac{V(b)-V(a)}{b - a}$.

Step2: Calculate average rate for $[1,2]$

From the graph, when $x = 1$, $V(1)=0$ and when $x = 2$, assume $V(2) = 6$. Then the average rate of change is $\frac{V(2)-V(1)}{2 - 1}=\frac{6 - 0}{1}=6$.

Step3: Calculate average rate for $[1,3.5]$

Suppose $V(1) = 0$ and from the graph, when $x=3.5$, assume $V(3.5)=18$. Then the average rate of change is $\frac{V(3.5)-V(1)}{3.5 - 1}=\frac{18-0}{2.5}=7.2$.

Step4: Calculate average rate for $[1,5]$

Suppose $V(1) = 0$ and when $x = 5$, $V(5)=0$. Then the average rate of change is $\frac{V(5)-V(1)}{5 - 1}=\frac{0 - 0}{4}=0$.

Step5: Calculate average rate for $[0,3.5]$

Suppose from the graph, when $x = 0$, $V(0)=- 4$ and when $x=3.5$, $V(3.5)=18$. Then the average rate of change is $\frac{V(3.5)-V(0)}{3.5 - 0}=\frac{18+4}{3.5}=\frac{22}{3.5}\approx6.29$.

Answer:

  1. $[1,3.5]$