We want to find which out of two cans of soup uses less material. This is the same as comparing the of the cans.
A can is a made up of three pieces — the top, the bottom, and the side. Both the top and the bottom are and the side is a . The rectangle's width is equal to the height of the cylinder and its length is equal to the of the circles.
The total surface area SA_\text{A} of a can is the combined of the circles A_\text{C} and the rectangle A_\text{R}.
\begin{gathered}
SA=2A_\text{C}+A_\text{R}
\end{gathered}
To find the surface area of Can B we will first need to find its so that we can use it to find the area of the circles. Let's start by doing this and then calculate the materials needed for the construction of each can one at a time.
To find the radius of Can B, we will use the fact that both cans have the same . We know that the radius r_\text{A} of Can A is half of the d_\text{A}.
\begin{gathered}
r_\text{A}=\dfrac{d_\text{A}}{2}=\dfrac{{\color{#0000FF}{3.5}}}{2}={\color{#0000FF}{1.75}} \text{ in.}
\end{gathered}
We now know that the radius of Can A is 1.75 inches, and that the height h_\text{A} of Can A is 3.6 inches. We can use this to find the volume of both cans using the formula for the .
V=\pi r_\text{A}^2 h_\text{A}
V=π ( 1.75)^2 ( 3.6)
V=π(3.0625)(3.6)
V=34.636059...
V≈ 34.6in.^3
V=\pi r_\text{B}^2 h_\text{B}
{\color{#FD9000}{34.6}}=\pi r_\text{B}^2 ({\color{#009600}{4.9}})
\dfrac{34.6}{4.9}=\pi r_\text{B}^2
\dfrac{34.6}{4.9\cdot\pi}=r_\text{B}^2
\sqrt{\dfrac{34.6}{4.9\cdot\pi}}=r_\text{B}
r_\text{B}=\sqrt{\dfrac{34.6}{4.9\cdot\pi}}
r_\text{B}=1.499219\ldots
r_\text{B}\approx {\color{#0000FF}{1.5}}
Material for Can A
Let's start by splitting up the surface area of the can into two circles and a rectangle.
The total surface area SA_\text{A} of the can is the combined area of the circles A_\text{CA} and the rectangle A_\text{RA}.
\begin{gathered}
SA_\text{A}=2A_\text{CA}+A_\text{RA}
\end{gathered}
We will start by calculating the area of each circle using the formula for the and the radius r_\text{A} of the circles.
A_\text{CA}=\pi r_\text{A}^2
A_\text{CA}=\pi ({\color{#0000FF}{1.75}})^2
A_\text{CA}=\pi (3.0625)
A_\text{CA}=9.621127\ldots
A_\text{CA}\approx {\color{#FF0000}{9.6}} \text{ in.}^2
C_\text{A}=2\pi r_\text{A}
C_\text{A}=2\pi ({\color{#0000FF}{1.75}})
C_\text{A}=10.995574\ldots
C_\text{A}\approx {\color{#FD9000}{11}} \text{ in.}
A_\text{RA}=\ell w
A_\text{RA}=C_\text{A} h_\text{A}
A_\text{RA}=({\color{#FD9000}{11}})({\color{#009600}{3.6}})
A_\text{RA}= {\color{#A800DD}{39.6}} \text{ in.}^2
Material for Can B
The method for calculating the amount of material needed for Can B is the same as for Can A. We start by splitting it into the top, bottom, and side.
The total surface area SA_\text{B} of the can is once again the combined area of the circles A_\text{CB} and the rectangle A_\text{RB}.
\begin{gathered}
SA_\text{B}=2A_\text{CB}+A_\text{RB}
\end{gathered}
Just like we did when calculating the surface area for Can A, we start by calculating the area of each circle.
A_\text{CB}=\pi r_\text{B}^2
A_\text{CB}=\pi ({\color{#0000FF}{1.5}})^2
A_\text{CB}=\pi (2.25)
A_\text{CB}=7.068583\ldots
A_\text{CB}\approx {\color{#FF0000}{7.1}} \text{ in.}^2
C_\text{B}=2\pi r_\text{A}
C_\text{B}=2\pi ({\color{#0000FF}{1.5}})
C_\text{B}=9.424777\ldots
C_\text{B}\approx {\color{#FD9000}{9.4}} \text{ in.}
A_\text{RB}=\ell w
A_\text{RB}=C_\text{B} h_\text{B}
A_\text{RB}=({\color{#FD9000}{9.4}})({\color{#009600}{4.9}})
A_\text{RB}=46.06
A_\text{RB}\approx {\color{#A800DD}{46}} \text{ in.}^2
