Answer to Question #266277 in Calculus for tim

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) =" the length of the rectangle at time t

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

The area of the rectangle at time t is:

"A = xy"

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

"dA \/ dt = ( dx \/ dt )( y ) + ( x )( dy \/ dt)\\\\\n\nx = 8\\\\\n\ny = 12\\\\\n\ndx \/ dt = 3\\\\\n\ndy \/ dt = 2\\\\\n\ndA \/ dt = ( 3 )( 12 ) + ( 8 )( 2 ) = 36 + 16 = 52\\ ft^{2 }\/ s\\\\"



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