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The side of the base of a square prism is increasing at a rate of 5 meters per second and the height of the prism is decreasing at a rate of 2 meters per second. At a certain instant, the base's side is 6 meters and the height is 7 meters. What is the rate of change of the volume of the prism at that instant (in cubic meters per second)? Choose 1 answer: (A) 348 (B) -492 (C) -348 (D) 492 The volume of a square prism with base side s and height h is s^(2)h.

The side of the base of a square prism is increasing at a rate of 55 meters per second and the height of the prism is decreasing at a rate of 22 meters per second.\newlineAt a certain instant, the base's side is 66 meters and the height is 77 meters.\newlineWhat is the rate of change of the volume of the prism at that instant (in cubic meters per second)?\newlineChoose 11 answer:\newline(A) 348348\newline(B) −492-492\newline(C) −348-348\newline(D) 492492\newlineThe volume of a square prism with base side s s and height h h is s2h s^{2} h .

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Q. The side of the base of a square prism is increasing at a rate of 55 meters per second and the height of the prism is decreasing at a rate of 22 meters per second.\newlineAt a certain instant, the base's side is 66 meters and the height is 77 meters.\newlineWhat is the rate of change of the volume of the prism at that instant (in cubic meters per second)?\newlineChoose 11 answer:\newline(A) 348348\newline(B) −492-492\newline(C) −348-348\newline(D) 492492\newlineThe volume of a square prism with base side s s and height h h is s2h s^{2} h .
  1. Volume Formula Explanation: The formula for the volume of a square prism is V=s2×hV = s^2 \times h, where ss is the side of the base and hh is the height.
  2. Rate of Change Derivation: To find the rate of change of the volume, we need to differentiate the volume with respect to time tt, so we get dVdt=2sdsdt⋅h+s2dhdt\frac{dV}{dt} = 2s\frac{ds}{dt} \cdot h + s^2\frac{dh}{dt}.
  3. Given Rates: Given dsdt=5 m/s\frac{ds}{dt} = 5 \, \text{m/s} (rate of change of the side) and dhdt=−2 m/s\frac{dh}{dt} = -2 \, \text{m/s} (rate of change of the height).
  4. Instant Values Substitution: At the instant when s=6 ms = 6\, \text{m} and h=7 mh = 7\, \text{m}, we plug these values into the differentiated volume formula: dVdt=2×6×5×7+62×(−2)\frac{dV}{dt} = 2 \times 6 \times 5 \times 7 + 6^2 \times (-2).
  5. Volume Rate Calculation: Calculate the rate of change of the volume: dVdt=2×6×5×7+62×(−2)=420+72×(−2)\frac{dV}{dt} = 2 \times 6 \times 5 \times 7 + 6^2 \times (-2) = 420 + 72 \times (-2).

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