The side of the base of a square pyramid is increasing at a rate of 6 meters per minute and the height of the pyramid is decreasing at a rate of 1 meter per minute. At a certain instant, the base's side is 3 meters and the height is 9 meters. What is the rate of change of the volume of the pyramid at that instant (in cubic meters per minute)? Choose 1 answer: (A) 111 (B) 105 (C) -105 (D) -111 The volume of a square pyramid with base side s and height h is (1)/(3)s^(2)h.

Question

The side of the base of a square pyramid is increasing at a rate of 6 meters per minute and the height of the pyramid is decreasing at a rate of 1 meter per minute. At a certain instant, the base's side is 3 meters and the height is 9 meters. What is the rate of change of the volume of the pyramid at that instant (in cubic meters per minute)? Choose 1 answer: (A)  111 (B)  105 (C) -105 (D) -111 The volume of a square pyramid with base side  s and height  h is  (1)/(3)s^(2)h.

Bytelearn · Accepted Answer

Volume Formula: The volume V of a square pyramid with base side s and height h is given by V = (1/3)s^2h.Differentiation with Respect to Time: To find the rate of change of the volume, we need to differentiate the volume formula with respect to time t, so we get dV/dt = (1/3)(2s(ds/dt)h + s^2(dh/dt)).Given Rates of Change: We know ds/dt = 6 meters/minute (rate of change of the side) and dh/dt = -1 meter/minute (rate of change of the height).Instant Values: At the instant we are considering, s = 3 meters and h = 9 meters.Plugging in Values: Plugging in the values, we get dV/dt = (1/3)(2*3*6*9 + 3^2*(-1)).Simplify Expression: Simplify the expression: dV/dt = (1/3)(2*3*6*9 + 9*(-1)).Calculate Expression: Calculate the expression: dV/dt = (1/3)(324 - 9).Further Simplification: Further simplification gives us dV/dt = (1/3)(315).Final Result: Finally, dV/dt = 105 cubic meters/minute.

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