An open-topped glass aquarium with a square base is designed to hold $32$ cubic feet of water. What is the minimum possible exterior surface area of the aquarium? $\newline$ square feet

Full solution

Q. An open-topped glass aquarium with a square base is designed to hold

32

cubic feet of water. What is the minimum possible exterior surface area of the aquarium?

\newline

square feet

Define Variables: Let $s$ be the side length of the square base and $h$ be the height of the aquarium. $\newline$ Volume of the aquarium $V = s^2 \times h$ .
Given Volume: Given volume $V = 32$ cubic feet. $\newline$ Substitute $32$ for $V$ in $V = s^2 \cdot h$ . $\newline$ $32 = s^2 \cdot h$ .
Substitute Volume: Since the base is square and we want the minimum surface area, the aquarium will be a cube (where all sides are equal). $\newline$ So, $s = h$ .
Base is a Cube: Substitute $s$ for $h$ in $32 = s^2 \cdot h$ .
$32 = s^2 \cdot s$ .
$32 = s^3$ .
Calculate Side Length: Solve for $s$ . $s = \sqrt[3]{32}$ . $s = 2 \times \sqrt[3]{2}$ .
Calculate Surface Area: Calculate the exterior surface area of the aquarium without the open top. $\newline$ Surface area $A = 4 \times s^2 + s^2$ ( $4$ sides plus the base).
Calculate Surface Area: Calculate the exterior surface area of the aquarium without the open top. $\newline$ Surface area $A = 4 \times s^2 + s^2$ ( $4$ sides plus the base).Substitute $2 \times \text{cube root of } 2$ for $s$ in $A = 4 \times s^2 + s^2$ . $\newline$ $A = 4 \times (2 \times \text{cube root of } 2)^2 + (2 \times \text{cube root of } 2)^2$ . $\newline$ $A = 4 \times 4 \times 2 + 4 \times 2$ . $\newline$ $A = 32 + 8$ . $\newline$ $A = 40$ square feet.