Question

A design team determines that a cost-efficient way of manufacturing cylindrical containers for their products is to have the volume, \(V\) in cubic centimetres, modelled by \(V(x)=9 \pi x^{3}+51 \pi x^{2}+88 \pi x+48 \pi,\) where \(x\) is an integer such that \(2 \leq x \leq 8 .\) The height, \(h,\) in centimetres, of each cylinder is a linear function given by \(h(x)=x+3\) a) Determine the quotient \(\frac{V(x)}{h(x)}\) and interpret this result. b) Use your answer in part a) to express the volume of a container in the form \(\pi r^{2} h\) c) What are the possible dimensions of the containers for the given values of \(x ?\)

Short answer

The quotient is \(9\pi x^2 + 24\pi x + 16\pi\), representing \(\pi r^2\). The volume is \((9x^2 + 24x + 16) \pi (x + 3)\). Possible dimensions vary based on values of \(x\) from 2 to 8.

Step-by-step solution

- Understand the Given Functions

Identify the functions provided in the problem. The volume function is given by \(V(x) = 9\pi x^3 + 51\pi x^2 + 88\pi x + 48\pi\), and the height function is given by \(h(x) = x + 3\).

- Find the Quotient \(\frac{V(x)}{h(x)}\)

To find the quotient, perform polynomial division of \(V(x)\) by \(h(x)\). Divide \(V(x) = 9\pi x^3 + 51\pi x^2 + 88\pi x + 48\pi\) by \(h(x) = x + 3\).

- Perform Polynomial Division

When dividing \(9\pi x^3 + 51\pi x^2 + 88\pi x + 48\pi\) by \(x + 3\), the quotient is found to be \(9\pi x^2 + 24\pi x + 16\pi\), with no remainder.

- Interpret the Quotient

The quotient \(\frac{V(x)}{h(x)} = 9\pi x^2 + 24\pi x + 16\pi\) represents the area of the base, \(\pi r^2\), where \(r\) is the radius of the cylinder.

- Express Volume in the Form \(\pi r^2 h\)

Since \(\frac{V(x)}{h(x)} = \pi r^2\), and from Step 3, \(\pi r^2 = 9\pi x^2 + 24\pi x + 16\pi\). This can be written as \(r^2 = 9x^2 + 24x + 16\). Therefore, \(V(x) = \pi r^2 h = (9x^2 + 24x + 16) (x + 3) \pi\).

- Determine Possible Dimensions

Given \(2 \leq x \leq 8\), calculate the radius and height for each integer value of \(x\). For example: \(x = 2\), \(h(2) = 5\), and \(r^2 = 128\). Thus, \(r = \sqrt{128} = 8\sqrt{2}\) cm. Repeat for \(x = 3, 4, 5, 6, 7, 8\).

Key concepts

Polynomial Division

Polynomial division is like long division but for polynomials. In this exercise, we divide one polynomial, the volume function, by a linear polynomial, the height function. This helps us find out how the volume is distributed in terms of height.

The volume function is given as \(V(x) = 9\pi x^3 + 51\pi x^2 + 88\pi x + 48\pi\). The height function is given as \(h(x) = x + 3\). We need to divide \(V(x)\) by \(h(x)\) to find the quotient without a remainder.

The quotient we get, \(9\pi x^2 + 24\pi x + 16\pi\), represents an important relationship between volume and height. This quotient gives us the base area of the cylinder, making things much simpler.

Cylindrical Volume

The volume of a cylinder is calculated using the formula \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. In our case, the volume function \(V(x)\) is given and depends on \(x\), and we have the height function \(h(x)\) as a linear equation.

First, we performed polynomial division to find the quotient, which represents \(\pi r^2\). From the division, we found \(\pi r^2 = 9\pi x^2 + 24\pi x + 16\pi\). This means that the base area of the cylinder is essentially \(9x^2 + 24x + 16\), and therefore, \(r^2 = 9x^2 + 24x + 16\).

With this, we can easily express the volume in the form \(\pi r^2 h\). We use the height function \(h(x) = x + 3\) we calculated.

Linear Functions

Linear functions are functions that form a straight line on a graph. They have the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. In this exercise, we are given a linear function for height, \(h(x) = x + 3\).

This indicates that the height of the cylinder increases by 1 unit for each unit increase in \(x\) and starts from a base height of 3 cm. This simple linear relationship helps us understand how the height affects the volume of the cylinder.

Once we determined the area of the base from polynomial division, we incorporated this linear function to get the complete volume expression as \(V(x) = (9x^2 + 24x + 16)(x + 3)\pi\).

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions using arithmetic operations and algebraic properties. In this problem, we performed several algebraic manipulations:

• First, we identified the given functions \(V(x)\) and \(h(x)\).
• Then, we divided \(V(x)\) by \(h(x)\) to obtain the quotient using polynomial division.
• Next, we interpreted the quotient to find the base area, represented by \(\pi r^2\).
• Finally, we expressed the volume in terms of \(\pi r^2 h\) and used the given range for \(x\) to find possible dimensions.

Breaking the problem down into these simpler steps helped us clearly understand each part and how to arrive at the correct solution.