Answer to Question #266277 in Calculus for tim

Question

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?

Answer 1

$x(t) =$ 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:

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