The surface area of a cube is increasing at a rate of 15 square meters per hour. At a certain instant, the surface area is 24 square meters. What is the rate of change of the volume of the cube at that instant (in cubic meters per hour)? Choose 1 answer: (A) (15)/(2) (B) (sqrt15)^(3) (C) (5)/(8) (D) 8

Find Volume of Cube: Now, let's find the volume of the cube. Volume = side^3 Volume = 2^3 Volume = 8 cubic metersRate of Change of Surface Area: Next, we need to find the rate of change of the surface area in terms of the side length. Since the surface area is 6 * side^2, the rate of change of the surface area is 6 * 2 * side * (d(side)/dt) Given that the surface area is increasing at 15 square meters per hour, we have: 6 * 2 * side * (d(side)/dt) = 15 12 * side * (d(side)/dt) = 15Solve for Rate of Change: Now we solve for (d(side)/dt), the rate of change of the side length. (d(side)/dt) = 15 / (12 * side) (d(side)/dt) = 15 / (12 * 2) (d(side)/dt) = 15 / 24 (d(side)/dt) = 5/8 meters per hourFind Rate of Change of Volume: Finally, we find the rate of change of the volume. The volume is side^3, so the rate of change of the volume is 3 * side^2 * (d(side)/dt) Rate of change of volume = 3 * 2^2 * (5/8) Rate of change of volume = 3 * 4 * (5/8) Rate of change of volume = 12 * (5/8) Rate of change of volume = 60/8 Rate of change of volume = 7.5 cubic meters per hour

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The surface area of a cube is increasing at a rate of 15 square meters per hour.
At a certain instant, the surface area is 24 square meters.
What is the rate of change of the volume of the cube at that instant (in cubic meters per hour)?
Choose 1 answer:
(A)
(15)/(2)
(B)
(sqrt15)^(3)
(C)
(5)/(8)
(D) 8

The surface area of a cube is increasing at a rate of $15$ square meters per hour. $\newline$ At a certain instant, the surface area is $24$ square meters. $\newline$ What is the rate of change of the volume of the cube at that instant (in cubic meters per hour)? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{15}{2}$ $\newline$ (B) $(\sqrt{15})^{3}$ $\newline$ (C) $\frac{5}{8}$ $\newline$ (D) $8$

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Math Problems

Grade 6

Area of quadrilaterals and triangles: word problems

Full solution

Q. The surface area of a cube is increasing at a rate of

15

square meters per hour.

\newline

At a certain instant, the surface area is

24

square meters.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

\frac{15}{2}

\newline

(B)

(\sqrt{15})^{3}

\newline

(C)

\frac{5}{8}

\newline

(D)

8

Find Volume of Cube: Now, let's find the volume of the cube. $\newline$ Volume = $\text{side}^3$ $\newline$ Volume = $2^3$ $\newline$ Volume = $8$ cubic meters
Rate of Change of Surface Area: Next, we need to find the rate of change of the surface area in terms of the side length. $\newline$ Since the surface area is $6 \times \text{side}^2$ , the rate of change of the surface area is $6 \times 2 \times \text{side} \times (\frac{d(\text{side})}{dt})$ $\newline$ Given that the surface area is increasing at $15$ square meters per hour, we have: $\newline$ $6 \times 2 \times \text{side} \times (\frac{d(\text{side})}{dt}) = 15$ $\newline$ $12 \times \text{side} \times (\frac{d(\text{side})}{dt}) = 15$
Solve for Rate of Change: Now we solve for $\frac{d(\text{side})}{dt}$ , the rate of change of the side length. $\frac{d(\text{side})}{dt} = \frac{15}{12 \times \text{side}}$ $\frac{d(\text{side})}{dt} = \frac{15}{12 \times 2}$ $\frac{d(\text{side})}{dt} = \frac{15}{24}$ $\frac{d(\text{side})}{dt} = \frac{5}{8}$ meters per hour
Find Rate of Change of Volume: Finally, we find the rate of change of the volume. $\newline$ The volume is $\text{side}^3$ , so the rate of change of the volume is $3 \times \text{side}^2 \times (\frac{d(\text{side})}{dt})$ $\newline$ Rate of change of volume = $3 \times 2^2 \times (\frac{5}{8})$ $\newline$ Rate of change of volume = $3 \times 4 \times (\frac{5}{8})$ $\newline$ Rate of change of volume = $12 \times (\frac{5}{8})$ $\newline$ Rate of change of volume = $\frac{60}{8}$ $\newline$ Rate of change of volume = $7.5$ cubic meters per hour