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anonymous
 5 years ago
at a certain instant the dimensions of a rectangular parallelepiped are 4,5,&6ft. and they are increasing respectively at the rates of 1,2,&3ft per second. at what rate is the volume changing?
anonymous
 5 years ago
at a certain instant the dimensions of a rectangular parallelepiped are 4,5,&6ft. and they are increasing respectively at the rates of 1,2,&3ft per second. at what rate is the volume changing?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0V = LWH You need the product rule to do this one. However, you must break it into two groups. dV/dt = d/dt[LW]*H+LW*d/dt[h] dV/dt = [dL/dt*W+L*dW/dt]H+LW*dH/dt dV/dt = WH*dL/dt + LH*dW/dt + LW*dH/dt L = 4; W = 5; H = 6; dL/dt = 1; dW/dt = 2; dH/dt = 3 dV/dt = 5*6*1+4*6*2+4*5*3 = 138 ft^3/s
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