Question

4. An ice cream shop wants to design a super straw to serve with its extra thick milkshakes that is double both the width and thickness of a standard straw.

A standard straw is 4 mm in diameter and 0.5 mm thick.
a. What is the cross-sectional (parallel to the base) area of the new straw (round to the nearest hundredth)?
b. If the new straw is 10 cm long, what is the maximum volume of milkshake that can be in the straw at one time (round to the nearest hundredth)?
c. A large milkshake is 32 fl.oz. (approximately 950 mL). If Corbin withdraws the full capacity of a straw 10 times a minute, what is the minimum amount of time that
it will take him to drink the milkshake (round to the nearest minute)?

Answer 1

A.
`\\ 0.50cm^(2)`; B.
`\\ 5.03cm^(3)`; C. 19min

Explanation:

First Step: Determine the size of the new straw

The size of the new straw is "double both the width and thickness of a standard straw". "A standard straw is 4mm in diameter and 0.5mm thick". So, the new straw is, then:

`\\ 2 * 4mm = 8mm` or, equivalently,
`\\ (1cm)/(10mm) * 8mm = 0.8cm` (Diameter).

`\\ 2 * 0.5mm = 1mm` or, equivalently,
`\\ (1cm)/(10mm) * 1mm = 0.1cm` (Thickness).

Second Step: Determine the cross-sectional area of the new straw

The cross-section area is "a section made by a plane cutting anything transversely, especially at right angles to the longest axis" Cross-section (2020), In Dictionary.com. Therefore, the area here is that of a circle:

`\\ A_{cross_(section)}= \pi * r^2`,

Where
`\\ \pi = 3.1415926535897932384626....`, and represents the ratio between a circle's circumference and its diameter, and r is the circle's radius or its diameter divided by 2.

Then, the area is:

`\\ A_{cross_(section)} = \pi * ((0.8cm)/(2))^2 = \pi * (0.4cm)^2= 0.502655cm^2`, rounded to the nearest hundredth:

`\\ A_{cross_(section)} = 0.50cm^2`

Third Step: Determine the maximum volume of milkshake that can be in the straw at one time

The new straw is 10cm long and its cross-section is, approximately,

`\\ A_{cross_(section)} = 0.502655cm^2`

Since the straw is a cylinder, and the volume of a cylinder is:

`\\ V_(straw) = \pi * r^2 * h = A_{cross_(section)} * h`, where h is the height of the straw. In this case is 10cm long.

So, the maximum volume of milkshake that can be in the straw at one time is the volume of the straw:

`\\ V_(straw) = A_{cross_(section)}*h = 0.502655cm^2 * 10cm = 5.02655cm^3`, rounded to the nearest hundredth:

`\\ V_(straw) = 5.03cm^3`

Fouth Step: Determine the minimum amount of time that it will take Corbin to drink the milkshake

A large milkshake is 950mL, and we know that:

`\\ 1000mL = 1000cm^3 = 1L`

So,

`\\ 1mL = 1cm^3`, therefore,
`\\ 950mL = 950cm^3`

"Corbin withdraws the full capacity of a straw 10 times a minute" or Corbin withdraws:

`\\ 10*V_(straw) = 10*5.02655cm^3 = 50.2655cm^3` every minute.

The large milkshake is
`\\ 950cm^3` . If Corbin withdraws
`\\ (50.2655cm^3)/(min)`, the minimum amount of time that it will take him to drink the milkshake is:

`\\ (950cm^3)/((50.2655cm^3)/(min)) = (950)/(50.2655)*(cm^3)/(cm^3)*min = 18.8996min`, since
`\\ (cm^3)/(cm^3) = 1`.

Rounded to the nearest minute, the minimum amount of time is 19min.