An ice cube is freezing in such a way that the side length [tex]$s$[/tex], in inches, is [tex]$s(t)=\frac{1}{2} t+4$[/tex], where [tex]$t$[/tex] is in hours. The surface area of the ice cube

Part A: Write an equation that gives the volume at [tex]$t$[/tex] hours after freezing begins. (2 points)

Part B: Find the surface area as a function of time, using composition, and determine its range. (4 points)

Part C: After how many hours will the surface area equal 294 square inches? Show all necessary calculations, and check for extraneous solutions.

Part A: Find the volume V ( t ) of the ice cube as a function of time t : V ( t ) = ( 2 1 ​ t + 4 ) 3
Part B: Find the surface area A ( t ) of the ice cube as a function of time t : A ( t ) = 6 ( 2 1 ​ t + 4 ) 2 and determine its range: [ 96 , ∞ )
Part C: Find the time t when the surface area A ( t ) is equal to 294 square inches: t = 6

Explanation

Problem Analysis The problem provides the side length of an ice cube as a function of time, s ( t ) = 2 1 ​ t + 4 , and asks us to find the volume and surface area as functions of time, as well as the time when the surface area equals 294 square inches.

Part A: Volume as a function of time Part A asks for the volume V as a function of time t . The volume of a cube is given by V = s 3 . Substituting s ( t ) = 2 1 ​ t + 4 into the volume formula, we get V ( t ) = ( 2 1 ​ t + 4 ) 3 .

Volume Equation Therefore, the volume of the ice cube at time t is given by: V ( t ) = ( 2 1 ​ t + 4 ) 3

Part B: Surface Area as a function of time Part B asks for the surface area A as a function of time t and its range. The surface area of a cube is given by A = 6 s 2 . Substituting s ( t ) = 2 1 ​ t + 4 into the surface area formula, we get A ( t ) = 6 ( 2 1 ​ t + 4 ) 2 .

Surface Area Equation So, the surface area of the ice cube at time t is given by: A ( t ) = 6 ( 2 1 ​ t + 4 ) 2

Range of Surface Area To find the range of A ( t ) , we consider that t represents time, so t ≥ 0 . When t = 0 , A ( 0 ) = 6 ( 2 1 ​ ( 0 ) + 4 ) 2 = 6 ( 4 ) 2 = 6 ( 16 ) = 96 . As t increases, A ( t ) also increases. Since there is no upper bound on t , the surface area can increase without bound. Therefore, the range of A ( t ) is [ 96 , ∞ ) .

Surface Area Range The range of the surface area function is: [ 96 , ∞ )

Part C: Time when Surface Area is 294 Part C asks to find the time t when the surface area A ( t ) is equal to 294 square inches. We set A ( t ) = 294 and solve for t : 6 ( 2 1 ​ t + 4 ) 2 = 294 . Dividing both sides by 6, we get ( 2 1 ​ t + 4 ) 2 = 49 . Taking the square root of both sides, we get 2 1 ​ t + 4 = ± 7 .

Solving for Time Solving for t in both cases: 2 1 ​ t = − 4 + 7 = 3 or 2 1 ​ t = − 4 − 7 = − 11 . So, t = 2 ( 3 ) = 6 or t = 2 ( − 11 ) = − 22 . Since t represents time, t must be non-negative. Therefore, t = 6 is the only valid solution. t = − 22 is an extraneous solution.

Final Time Therefore, the surface area will equal 294 square inches after 6 hours. t = 6


Examples
Imagine you're designing ice sculptures for an event. Knowing how the volume and surface area of the ice change over time, as demonstrated in this problem, helps you predict how long your sculptures will last and how quickly they'll melt. This understanding is crucial for planning and executing successful ice sculpture displays.

Answered by GinnyAnswer | 2025-07-08

The volume of the ice cube as a function of time is given by V ( t ) = ( 2 1 ​ t + 4 ) 3 . The surface area is given by A ( t ) = 6 ( 2 1 ​ t + 4 ) 2 with a range of [ 96 , ∞ ) . The surface area equals 294 square inches at t = 6 hours.
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Answered by Anonymous | 2025-07-21