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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) 8 (B) (15)/(2) (C) (sqrt15)^(3) (D) (5)/(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) 88\newline(B) 152 \frac{15}{2} \newline(C) (15)3 (\sqrt{15})^{3} \newline(D) 58 \frac{5}{8}

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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) 88\newline(B) 152 \frac{15}{2} \newline(C) (15)3 (\sqrt{15})^{3} \newline(D) 58 \frac{5}{8}
  1. Surface Area Calculation: The surface area of a cube is 66 times one of its side's square, so one side's square is 24 m26\frac{24 \, \text{m}^2}{6}.\newlineCalculate the length of one side.\newlineSide length = 24 m26=4 m2=2 m\sqrt{\frac{24 \, \text{m}^2}{6}} = \sqrt{4 \, \text{m}^2} = 2 \, \text{m}.
  2. Volume Calculation: The volume of a cube is side length cubed, V=s3V = s^3. Calculate the volume with the side length. Volume = 2 m×2 m×2 m=8 m32 \, \text{m} \times 2 \, \text{m} \times 2 \, \text{m} = 8 \, \text{m}^3.
  3. Surface Area Rate of Change: The surface area is increasing at a rate of 15 m2/hour15 \, \text{m}^2/\text{hour}. Since the surface area is 6×side length26 \times \text{side length}^2, d(SA)dt=6×2×side length×d(side length)dt\frac{d(SA)}{dt} = 6 \times 2 \times \text{side length} \times \frac{d(\text{side length})}{dt}.
  4. Side Length Rate of Change: Solve for d(side length)dt\frac{d(\text{side length})}{dt}.15 m2/hour=6×2×2 m×d(side length)dt15 \, \text{m}^2/\text{hour} = 6 \times 2 \times 2 \, \text{m} \times \frac{d(\text{side length})}{dt}.d(side length)dt=15 m2/hour(6×2×2 m)=15 m2/hour24 m=58 m/hour\frac{d(\text{side length})}{dt} = \frac{15 \, \text{m}^2/\text{hour}}{(6 \times 2 \times 2 \, \text{m})} = \frac{15 \, \text{m}^2/\text{hour}}{24 \, \text{m}} = \frac{5}{8} \, \text{m/hour}.
  5. Volume Rate of Change: The rate of change of the volume is given by dVdt=3×side length2×d(side length)dt\frac{dV}{dt} = 3 \times \text{side length}^2 \times \frac{d(\text{side length})}{dt}. Calculate dVdt\frac{dV}{dt} using the side length and d(side length)dt\frac{d(\text{side length})}{dt}. dVdt=3×(2 m)2×(58 m/hour)=3×4 m2×58 m/hour=152 m3/hour.\frac{dV}{dt} = 3 \times (2 \, \text{m})^2 \times \left(\frac{5}{8} \, \text{m/hour}\right) = 3 \times 4 \, \text{m}^2 \times \frac{5}{8} \, \text{m/hour} = \frac{15}{2} \, \text{m}^3/\text{hour}.

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