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

Ihad to look for the figure and here is my answer:

based on the given circle d with angles adc and arc ac, here is how we are going to look for the measure of angle abc:
arc ac=angle adc
8x-8=7x+2
8x-7x=2+8
x=10
< adc=7(10)+2=72
< abc=72/2=36
so the answer for this is 36 degrees.