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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 1515 square meters per hour.\newlineAt a certain instant, the surface area is 2424 square meters.\newlineWhat is the rate of change of the volume of the cube at that instant (in cubic meters per hour)?\newlineChoose 11 answer:\newline(A) 152 \frac{15}{2} \newline(B) (15)3 (\sqrt{15})^{3} \newline(C) 58 \frac{5}{8} \newline(D) 88

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Q. The surface area of a cube is increasing at a rate of 1515 square meters per hour.\newlineAt a certain instant, the surface area is 2424 square meters.\newlineWhat is the rate of change of the volume of the cube at that instant (in cubic meters per hour)?\newlineChoose 11 answer:\newline(A) 152 \frac{15}{2} \newline(B) (15)3 (\sqrt{15})^{3} \newline(C) 58 \frac{5}{8} \newline(D) 88
  1. Find Volume of Cube: Now, let's find the volume of the cube.\newlineVolume = side3\text{side}^3\newlineVolume = 232^3\newlineVolume = 88 cubic meters
  2. 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.\newlineSince the surface area is 6×side26 \times \text{side}^2, the rate of change of the surface area is 6×2×side×(d(side)dt)6 \times 2 \times \text{side} \times (\frac{d(\text{side})}{dt})\newlineGiven that the surface area is increasing at 1515 square meters per hour, we have:\newline6×2×side×(d(side)dt)=156 \times 2 \times \text{side} \times (\frac{d(\text{side})}{dt}) = 15\newline12×side×(d(side)dt)=1512 \times \text{side} \times (\frac{d(\text{side})}{dt}) = 15
  3. Solve for Rate of Change: Now we solve for d(side)dt\frac{d(\text{side})}{dt}, the rate of change of the side length.d(side)dt=1512×side\frac{d(\text{side})}{dt} = \frac{15}{12 \times \text{side}}d(side)dt=1512×2\frac{d(\text{side})}{dt} = \frac{15}{12 \times 2}d(side)dt=1524\frac{d(\text{side})}{dt} = \frac{15}{24}d(side)dt=58\frac{d(\text{side})}{dt} = \frac{5}{8} meters per hour
  4. Find Rate of Change of Volume: Finally, we find the rate of change of the volume.\newlineThe volume is side3\text{side}^3, so the rate of change of the volume is 3×side2×(d(side)dt)3 \times \text{side}^2 \times (\frac{d(\text{side})}{dt})\newlineRate of change of volume = 3×22×(58)3 \times 2^2 \times (\frac{5}{8})\newlineRate of change of volume = 3×4×(58)3 \times 4 \times (\frac{5}{8})\newlineRate of change of volume = 12×(58)12 \times (\frac{5}{8})\newlineRate of change of volume = 608\frac{60}{8}\newlineRate of change of volume = 7.57.5 cubic meters per hour

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