Question #266277

At a certain instant the dimensions of the rectangle are 8 and 12 feet, and they are increasing at the rates 3 and 2 feet per second, respectively. How fast is the area changing?


1
Expert's answer
2021-11-15T20:07:15-0500

x(t)=x(t) = the length of the rectangle at time t

y(t)=y(t) = the width of the rectangle at time t

The area of the rectangle at time t is:

A=xyA = xy

The rate at which the area of the rectangle changes is:

dA/dt=(dx/dt)(y)+(x)(dy/dt)x=8y=12dx/dt=3dy/dt=2dA/dt=(3)(12)+(8)(2)=36+16=52 ft2/sdA / dt = ( dx / dt )( y ) + ( x )( dy / dt)\\ x = 8\\ y = 12\\ dx / dt = 3\\ dy / dt = 2\\ dA / dt = ( 3 )( 12 ) + ( 8 )( 2 ) = 36 + 16 = 52\ ft^{2 }/ s\\



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