Math / Algebra

QuestionAn ice cube is freezing in such a way that the side length $s$ , in inches, is $s(t)=\frac{1}{2} t+4$ , where $t$ is in hours. The surface area of the ice cube is the function $A(s)=6 s^{2}$ . Part A: Write an equation that gives the volume at $t$ 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. (4 points)

Studdy Solution

STEP 1

1. The side length $s$ of the ice cube is given by $s(t) = \frac{1}{2} t + 4$ .
2. The surface area of the ice cube is given by $A(s) = 6s^2$ .
3. The volume of a cube is given by $V = s^3$ .

STEP 2

1. Part A: Derive the equation for the volume of the ice cube as a function of time $t$ .
2. Part B: Find the surface area as a function of time using composition and determine its range.
3. Part C: Solve for $t$ when the surface area equals 294 square inches and check for extraneous solutions.

STEP 3

To find the volume as a function of time, use the formula for the volume of a cube, $V = s^3$ .
Substitute $s(t) = \frac{1}{2} t + 4$ into the volume formula:
$V(t) = \left( \frac{1}{2} t + 4 \right)^3$

STEP 4

To find the surface area as a function of time, substitute $s(t) = \frac{1}{2} t + 4$ into the surface area formula $A(s) = 6s^2$ .
$A(t) = 6 \left( \frac{1}{2} t + 4 \right)^2$

STEP 5

Expand the expression for $A(t)$ :
$A(t) = 6 \left( \left( \frac{1}{2} t + 4 \right)^2 \right) = 6 \left( \frac{1}{4}t^2 + 4t + 16 \right)$
$A(t) = \frac{3}{2}t^2 + 24t + 96$

STEP 6

Determine the range of $A(t)$ . Since $A(t)$ is a quadratic function opening upwards, the minimum value occurs at $t = 0$ .
Calculate $A(0)$ :
$A(0) = \frac{3}{2}(0)^2 + 24(0) + 96 = 96$
Thus, the range of $A(t)$ is $[96, \infty)$ .

STEP 7

To find when the surface area equals 294 square inches, solve the equation:
$\frac{3}{2}t^2 + 24t + 96 = 294$
Simplify and solve for $t$ :
$\frac{3}{2}t^2 + 24t + 96 - 294 = 0$
$\frac{3}{2}t^2 + 24t - 198 = 0$
Multiply through by 2 to clear the fraction:
$3t^2 + 48t - 396 = 0$

STEP 8

Use the quadratic formula to solve for $t$ :
$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 3$ , $b = 48$ , and $c = -396$ .
Calculate the discriminant:
$b^2 - 4ac = 48^2 - 4 \times 3 \times (-396)$
$= 2304 + 4752 = 7056$
Calculate $t$ :
$t = \frac{-48 \pm \sqrt{7056}}{6}$
$\sqrt{7056} = 84$
$t = \frac{-48 \pm 84}{6}$

STEP 9

Calculate the two possible solutions for $t$ :
$t = \frac{-48 + 84}{6} = \frac{36}{6} = 6$
$t = \frac{-48 - 84}{6} = \frac{-132}{6} = -22$
Since $t$ must be non-negative, the valid solution is $t = 6$ .
The solution for Part A is: $V(t) = \left( \frac{1}{2} t + 4 \right)^3$
The solution for Part B is: $A(t) = \frac{3}{2}t^2 + 24t + 96$ Range: $[96, \infty)$
The solution for Part C is: $t = 6$

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Studdy Solution

STEP 1

1. The side length $s$ of the ice cube is given by $s(t) = \frac{1}{2} t + 4$ .
2. The surface area of the ice cube is given by $A(s) = 6s^2$ .
3. The volume of a cube is given by $V = s^3$ .

STEP 2

1. Part A: Derive the equation for the volume of the ice cube as a function of time $t$ .
2. Part B: Find the surface area as a function of time using composition and determine its range.
3. Part C: Solve for $t$ when the surface area equals 294 square inches and check for extraneous solutions.

STEP 3

To find the volume as a function of time, use the formula for the volume of a cube, $V = s^3$ .
Substitute $s(t) = \frac{1}{2} t + 4$ into the volume formula:
$V(t) = \left( \frac{1}{2} t + 4 \right)^3$

STEP 4

To find the surface area as a function of time, substitute $s(t) = \frac{1}{2} t + 4$ into the surface area formula $A(s) = 6s^2$ .
$A(t) = 6 \left( \frac{1}{2} t + 4 \right)^2$

STEP 5

Expand the expression for $A(t)$ :
$A(t) = 6 \left( \left( \frac{1}{2} t + 4 \right)^2 \right) = 6 \left( \frac{1}{4}t^2 + 4t + 16 \right)$
$A(t) = \frac{3}{2}t^2 + 24t + 96$

STEP 6

Determine the range of $A(t)$ . Since $A(t)$ is a quadratic function opening upwards, the minimum value occurs at $t = 0$ .
Calculate $A(0)$ :
$A(0) = \frac{3}{2}(0)^2 + 24(0) + 96 = 96$
Thus, the range of $A(t)$ is $[96, \infty)$ .

STEP 7

To find when the surface area equals 294 square inches, solve the equation:
$\frac{3}{2}t^2 + 24t + 96 = 294$
Simplify and solve for $t$ :
$\frac{3}{2}t^2 + 24t + 96 - 294 = 0$
$\frac{3}{2}t^2 + 24t - 198 = 0$
Multiply through by 2 to clear the fraction:
$3t^2 + 48t - 396 = 0$

STEP 8

STEP 9

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Studdy Solution

STEP 1

1. The side length s s s of the ice cube is given by s(t)=12t+4 s(t) = \frac{1}{2} t + 4 s(t)=21​t+4.2. The surface area of the ice cube is given by A(s)=6s2 A(s) = 6s^2 A(s)=6s2.3. The volume of a cube is given by V=s3 V = s^3 V=s3.

STEP 2

1. Part A: Derive the equation for the volume of the ice cube as a function of time t t t.2. Part B: Find the surface area as a function of time using composition and determine its range.3. Part C: Solve for t t t when the surface area equals 294 square inches and check for extraneous solutions.

STEP 3

To find the volume as a function of time, use the formula for the volume of a cube, V=s3 V = s^3 V=s3.Substitute s(t)=12t+4 s(t) = \frac{1}{2} t + 4 s(t)=21​t+4 into the volume formula:V(t)=(12t+4)3 V(t) = \left( \frac{1}{2} t + 4 \right)^3 V(t)=(21​t+4)3

STEP 4

To find the surface area as a function of time, substitute s(t)=12t+4 s(t) = \frac{1}{2} t + 4 s(t)=21​t+4 into the surface area formula A(s)=6s2 A(s) = 6s^2 A(s)=6s2.A(t)=6(12t+4)2 A(t) = 6 \left( \frac{1}{2} t + 4 \right)^2 A(t)=6(21​t+4)2

STEP 5

Expand the expression for A(t) A(t) A(t):A(t)=6((12t+4)2)=6(14t2+4t+16) A(t) = 6 \left( \left( \frac{1}{2} t + 4 \right)^2 \right) = 6 \left( \frac{1}{4}t^2 + 4t + 16 \right) A(t)=6((21​t+4)2)=6(41​t2+4t+16)A(t)=32t2+24t+96 A(t) = \frac{3}{2}t^2 + 24t + 96 A(t)=23​t2+24t+96

STEP 6

STEP 7

STEP 8

STEP 9

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1. The side length $s$ of the ice cube is given by $s(t) = \frac{1}{2} t + 4$ .
2. The surface area of the ice cube is given by $A(s) = 6s^2$ .
3. The volume of a cube is given by $V = s^3$ .

1. Part A: Derive the equation for the volume of the ice cube as a function of time $t$ .
2. Part B: Find the surface area as a function of time using composition and determine its range.
3. Part C: Solve for $t$ when the surface area equals 294 square inches and check for extraneous solutions.

To find the volume as a function of time, use the formula for the volume of a cube, $V = s^3$ .
Substitute $s(t) = \frac{1}{2} t + 4$ into the volume formula:
$V(t) = \left( \frac{1}{2} t + 4 \right)^3$

To find the surface area as a function of time, substitute $s(t) = \frac{1}{2} t + 4$ into the surface area formula $A(s) = 6s^2$ .
$A(t) = 6 \left( \frac{1}{2} t + 4 \right)^2$

Expand the expression for $A(t)$ :
$A(t) = 6 \left( \left( \frac{1}{2} t + 4 \right)^2 \right) = 6 \left( \frac{1}{4}t^2 + 4t + 16 \right)$
$A(t) = \frac{3}{2}t^2 + 24t + 96$