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The side of a cube is decreasing at a rate of 9 millimeters per minute.
At a certain instant, the side is 19 millimeters.
What is the rate of change of the volume of the cube at that instant (in cubic millimeters per minute)?
Choose 1 answer:
(A) -3249
(B) -6859
(C) -9747
(D) -729

The side of a cube is decreasing at a rate of $9$ millimeters per minute. $\newline$ At a certain instant, the side is $19$ millimeters. $\newline$ What is the rate of change of the volume of the cube at that instant (in cubic millimeters per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $-3249$ $\newline$ (B) $-6859$ $\newline$ (C) $-9747$ $\newline$ (D) $-729$

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Calculate unit rates with fractions

Full solution

Q. The side of a cube is decreasing at a rate of

9

millimeters per minute.

\newline

At a certain instant, the side is

19

millimeters.

\newline

What is the rate of change of the volume of the cube at that instant (in cubic millimeters per minute)?

\newline

Choose

1

answer:

\newline

(A)

-3249

\newline

(B)

-6859

\newline

(C)

-9747

\newline

(D)

-729

Volume Formula Derivation: The volume of a cube is given by the formula $V = s^3$ , where $s$ is the side length of the cube. To find the rate of change of the volume, we need to differentiate the volume with respect to time.
Rate of Change Calculation: So, $\frac{dV}{dt} = 3s^2 \cdot \frac{ds}{dt}$ . Here, $\frac{ds}{dt}$ is the rate of change of the side length, which is $-9\,\text{mm/min}$ (negative because the side is decreasing).
Substitute Values: Now we plug in the values: $s = 19 \, \text{mm}$ and $\frac{ds}{dt} = -9 \, \text{mm/min}$ . So, $\frac{dV}{dt} = 3 \times (19)^2 \times (-9)$ .
Final Calculation: Calculating that out, $\frac{dV}{dt} = 3 \times 361 \times (-9) = 3 \times (-3249) = -9747$ cubic millimeters per minute.

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