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an ice cube is freesinf in such a way thag ghe side length s,in inches, is s(t)=1/2t+2 where t is in hours. The surface area of the ice cube is the function A(s)=6s^2. Part A: Write an equstion that gives the volume at t hours after freezing begins. Part B: Find the surface area as the function kf time, using composition, and determjne its range.Part C: After how many hours will the surface area equak 216 squRs inches? Show all neccesady calculations.

An Ice Cube Is Freesinf In Such A Way Thag Ghe Side Length Sin Inches Is St12t2 Where T Is In Hours The Surface Area Of The Ice Cube Is The Function As6s2 Part class=

Sagot :

Part A

we have the function

[tex]S(t)=\frac{1}{2}t+2[/tex]

Remember that

The volume of a cube is given by the formula

[tex]V=S^3[/tex]

substitute the function S(t) in the formula of volume

[tex]V(t)=(\frac{1}{2}t+2)^3[/tex]

Part B

Find the surface area as the function of time

we have

[tex]SA(S)=6S^2[/tex]

Find out (SAoS)(t)=SA(S(t))

so

[tex]SA(S(t))=6(\frac{1}{2}t+2)^2[/tex]

Part C

we have

SA=216 in2

using the function of Part B

[tex]\begin{gathered} SA(t)=6(\frac{1}{2}t+2)^2 \\ \\ 6(\frac{1}{2}t+2)^2=216 \end{gathered}[/tex]

Solve for t

[tex]\begin{gathered} (\frac{1}{2}t+2)^2=\frac{216}{6} \\ \\ (\frac{1}{2}t+2)^2=36 \\ take\text{ square root on both sides} \\ \\ \frac{1}{2}t+2=\pm6 \\ \\ \frac{1}{2}t=-2\pm6 \\ \\ t=-4\pm12 \end{gathered}[/tex]

The values of t are

t=8 hours and t=-16 hours ( is not a solution because is a negative number)

therefore

The answer is 8 hours

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