Question

PLEASE HELP!!
An ice cube is freezing in such a way that the side length s, in inches, is s of t equals one half times t plus 4 comma where t is in hours. The surface area of the ice cube is the function A(s) = 6s2.Part B: Find the surface area as a function of time, using composition, and determine its range. (4 points)

Answer 1

Answer:

A(t) = 3/2t² + 24t + 96

Range = (96, ∞)

Explanation:

The equation for the side length of the cube s is given by

$s(t)=\frac{1}{2}t+4$

Where t is the number of hours. In the same way, the equation for the surface area is:

$A(s)=6s^2$

Then, the surface area as a function of time will be the composite function A(s(t)). So, replacing s by the equation of s(t), we get:

$\begin{gathered} A(s(t))=6s(t)^2 \\ A(s(t))=6(\frac{1}{2}t+4)^2 \\ A(t_{})=6(\frac{1}{4}t^2+2(\frac{1}{2}t)(4)+4^2) \\ A(t)=6(\frac{1}{4}t^2+4t+16) \\ A(t)=6(\frac{1}{4}t^2)+6(4t)+6(16) \\ A(t)=\frac{3}{2}t^2+24t+96 \end{gathered}$

Then, the range is the set of all the possible values that A(t) can take. Since t takes values greater than or equal to 0, the minimum value that A(t) will take is 96 because:

A(0) = 3/2(0)² + 24(0) + 96 = 96

Therefore, the range for the surface area will be (96, ∞)