A rectangle has width w inches and height h, where the width is twice the height. Both w and h are functions of time, t, measured in seconds. If A represents the area of the rectangle, what is the rate of change of A with respect to t at the instant where the width is 4 inches and the height is increasing at the rate of 2 inches per second?

dA/dt = 16 square inches per second

Step-by-step explanation:

Width of rectangle is w

Height of rectangle is h

width = twice height

w = 2h

Area = wh = (2h)*h

A = 2h^2

Differentiating the equation with respect to time

dA/dt = 2+2h dh/dt

dA/dt = 4h dh/dt

According to the given situation when width is 4 inches

h = w/2

h = 2

Rate of change of A is  

dA/dt = wh dh/dt

dA/dt = 4(2)(2)

dA/dt = 16 square inches per second

6 hours 48 min

step-by-step explanation:

68÷5=13.6

13.6×30=408min=6 hours 48 min.

hope to and hope that my answer is brainliest.(follow me )

stops on an even number.