Question

Henry wants to make spaghetti and meatballs for his significant other. As he searches thekitchen, he comes up with a very small sauce pan. It’s only 3 inches deep and 4 inches indiameter. After filling the pan ½ way with sauce, he hopes he will be able to get 2 servings* ofmeatballs into the pan before it spills over. Is he able to? Use mathematical calculations tosupport your conclusion. *Assume meatballs are approximately 1 inch in diameter and eachserving is 5 meatballs. Ok

Answer 1

Given the dimension of the sauce pan

$\begin{gathered} \text{height, h=3inches} \\ \text{diameter,d=4 inches} \end{gathered}$

The sauce pan is cylindrical in shape, the volume of a cylinder is given as

$\begin{gathered} V=\pi r^2h \\ r=\frac{d}{2}=\frac{4}{2}=2in \\ h=3in \\ \pi=3.14 \end{gathered}$

The volume of the sauce pan will be

$\begin{gathered} V_{sp}=3.14\times2^2\times4=16\pi \\ =50.24in^3 \end{gathered}$

If the pan is filled half way, then the volume of the sauce is

$\begin{gathered} \frac{V_{sp}}{2}=\frac{50.24}{2}_{} \\ =25.12in^3 \end{gathered}$

The meatball is spherical in shape, it has a diameter of 1 inch, the volume of a sphere is

$\begin{gathered} V_{mb}=\frac{4}{3}\pi r^3 \\ d=1in \\ r=\frac{1}{2}=0.5in \\ \pi=3.14 \end{gathered}$

Hence, the volume is given by,

$V_{mb}=\frac{4}{3}\pi(\frac{1}{2})^3=\frac{4\pi}{3\times8}=\frac{3.14}{6}=0.52in^3$

So, 5 meat balls are in one serving, therefore the volume for one serving will be

$5\times V_{mb}=5\times0.52=2.6in^3$

The volume of two servings of meat ball will be

$2.6+2.6=5.2in^3$

Comparing the volume of sauce (25 cubic inches) he made in the pan and the volume of 2 meatballs (5.2 cubic inches) servings, we can confidently say he will be able to get two servings of meatballs from the sauce pan

Hence, he is able to get 2 servings of meatballs