A tube is being stretched while maintaining its cylindrical shape. The height is increasing at the rate of 2 millimeters per second. At the instant that the radius of the tube is 6 millimeters, the volume is increasing at the rate of 96π cubic millimeters per second. Which of the following statements about the surface area of the tube is true at this instant? (The volume V of a cylinder with radius r and height h is V=πr2h. The surface area S of a cylinder, not including the top and bottom of the cylinder, is S=2πrh.)

V = πr²h

S=2πrh

h= 2/3 millimeters

new height= 8/3 millimeters

S= 8π  millimeters square

New surface area = 32π  millimeters square

Step-by-step explanation:

The volume of the cylinder is given by V = πr²h   where r is the radius and h is the height.

The surface area of the cylinder is given by  S= 2πrh + 2πr²

Where πr² gives the area of the base and 2πr² gives the area of the top and bottom surfaces. The surface area S of a cylinder, not including the top and bottom of the cylinder, is therefore S=2πrh.

V = πr²h

96π= π (6*6) (h+2)

96 = 36 (h+2)

96/36= h+2

h= 96/36-2

h= 96-72/36

h= 24/36

h= 4/6

h= 2/3 millimeters

New height

h + 2= 2/3 + 2

= 2+6/3= 8/3 millimeters

Now S =2πrh

S = 2π(6) (2/3)

S= 8π  millimeters square

New Surface area

S = 2π(6) (8/3)

S= 32π  millimeters square

I also don't know