Chapter 1.2, Problem 55PS

Solution

a.

To find: The moisture content that maximizes the popping volume for hot-air popping and find the maximum volume.

The moisture content $x=14.06$ maximizes the popping volume and the maximum popping volume is $\text{55}\text{.5}$ cubic centimeter per gram.

Given information:

The equation of popping volume for hot-air popping is $y=-0.761(x-5.52)(x-22.6)$ .

Concept used:

The x-coordinate of the vertex of a quadratic function of the form $y=a\left(x-p\right)\left(x-q\right)$ is given by $x=\frac{p+q}{2}$ .

The vertex of the quadratic function represents the maximum/minimum point of the function.

If $a$ positive, the vertex is the minimum point and if $a$ is negative, the vertex is the maximum point of the function.

Calculation:

For the given function $y=-0.761(x-5.52)(x-22.6)$ , $p=5.52$ and $q=22.6$ .

Then,

  $\begin{array}{c}x=\frac{5.52+22.6}{2}\\ =\frac{28.12}{2}\\ =14.06\end{array}$

Therefore, the moisture content $x=14.06$ maximizes the popping volume.

Now, substitute $x=14.06$ in the function to find the maximum popping volume.

  $\begin{array}{c}y=-0.761(14.06-5.52)(14.06-22.6)\\ =-0.761\cdot 8.54\left(14.06-22.6\right)\\ =0.761\cdot 8.54\cdot 8.54\\ =\text{55}\text{.5009476}\end{array}$

Therefore, the maximum popping volume is $\text{55}\text{.5}$ cubic centimeter per gram.

Conclusion:

The moisture content $x=14.06$ maximizes the popping volume and the maximum popping volume is $\text{55}\text{.5}$ cubic centimeter per gram.

b.

To find: The moisture content that maximizes the popping volume for hot-oil popping and find the maximum volume.

The moisture content $x=13.575$ maximizes the popping volume and the maximum popping volume is $44.11$ cubic centimeter per gram.

Given information:

The equation of popping volume for hot-oil popping is $y=-0.652(x-5.35)(x-21.8)$ .

Concept used:

The x-coordinate of the vertex of a quadratic function of the form $y=a\left(x-p\right)\left(x-q\right)$ is given by $x=\frac{p+q}{2}$ .

The vertex of the quadratic function represents the maximum/minimum point of the function.

If $a$ positive, the vertex is the minimum point and if $a$ is negative, the vertex is the maximum point of the function.

Calculation:

For the given function $y=-0.652(x-5.35)(x-21.8)$ , $p=5.35$ and $q=21.8$ .

Then,

  $\begin{array}{c}x=\frac{5.35+21.8}{2}\\ =\frac{27.15}{2}\\ =13.575\end{array}$

Therefore, the moisture content $x=13.575$ maximizes the popping volume.

Now, substitute $x=13.575$ in the function to find the maximum popping volume.

  $\begin{array}{c}y=-0.652\left(13.575-5.35\right)\left(13.575-21.8\right)\\ =\text{44}\text{.1082075}\\ \approx \text{44}\text{.11}\end{array}$

Therefore, the maximum popping volume is $44.11$ cubic centimeter per gram.

Conclusion:

The moisture content $x=13.575$ maximizes the popping volume and the maximum popping volume is $44.11$ cubic centimeter per gram.

c.

To graph: Graph the functions on the same coordinate plane and find the domain and range of each function.

  

The domain of the function $y=-0.761(x-5.52)(x-22.6)$ is $5.52\le x\le 22.6$ , and the range is $0\le y\le 55.5$ .

The domain of the function $y=-0.652(x-5.35)(x-21.8)$ is $5.35\le x\le 21.8$ and the range is $0\le y\le 44.11$ .

Given information:

The equation of popping volume for hot-oil popping is $y=-0.652(x-5.35)(x-21.8)$ .

Concept used:

The x-intercept of the function of the form $y=a\left(x-p\right)\left(x-q\right)$ are $x=p$ and $x=q$ .

The vertex of the quadratic function represents the maximum/minimum point of the function.

If $a$ positive, the vertex is the minimum point and if $a$ is negative, the vertex is the maximum point of the function.

Graph:

Use a graphing calculator to graph the functions on the same coordinate plane.

  

Calculation:

The x-intercept of the function $y=-0.761(x-5.52)(x-22.6)$ are $x=5.52$ and $x=22.6$ .

Since volume cannot be negative, the graph only exists for non-negative y-values. That means, the graph exists between the x-intercepts of the function.

So, the domain of the function $y=-0.761(x-5.52)(x-22.6)$ is $5.52\le x\le 22.6$ .

The minimum value of the function on its domain is 0 and the maximum value is 55.5. So, the range of the function is $0\le y\le 55.5$ .

Similarly, the domain of the function $y=-0.652(x-5.35)(x-21.8)$ is $5.35\le x\le 21.8$ and the range is $0\le y\le 44.11$ .

Conclusion:

The domain of the function $y=-0.761(x-5.52)(x-22.6)$ is $5.52\le x\le 22.6$ , and the range is $0\le y\le 55.5$ .

The domain of the function $y=-0.652(x-5.35)(x-21.8)$ is $5.35\le x\le 21.8$ and the range is $0\le y\le 44.11$ .