Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rates of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now, with the rate equation known, we can solve for the rate of change of the surface area with what we know about the cube, namely that its sides have a length of 11 and a rate of growth of 30:

\(\displaystyle \frac{dA}{dt}=12(11)(30)=3960\)

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rates of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now, with the rate equation known, we can solve for the rate of change of the surface area with what we know about the cube, namely that its sides have a length of 10 and a rate of growth of 31:

\(\displaystyle \frac{dA}{dt}=12(10)(31)=3720\)

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rates of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now, with the rate equation known, we can solve for the rate of change of the surface area with what we know about the cube, namely that its sides have a length of 9 and a rate of growth of 32:

\(\displaystyle \frac{dA}{dt}=12(9)(32)=3456\)

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rates of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now, with the rate equation known, we can solve for the rate of change of the surface area with what we know about the cube, namely that its sides have a length of 8 and a rate of growth of 33:

\(\displaystyle \frac{dA}{dt}=12(8)(33)=3168\)

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 7 and a rate of growth of 34:

\(\displaystyle \frac{dA}{dt}=12(7)(34)=2856\)

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 6 and a rate of growth of 35:

\(\displaystyle \frac{dA}{dt}=12(6)(35)=2520\)

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 5 and a rate of growth of 36:

\(\displaystyle \frac{dA}{dt}=12(5)(36)=2160\)

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 4 and a rate of growth of 37:

\(\displaystyle \frac{dA}{dt}=12(4)(37)=1776\)

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 3 and a rate of growth of 38:

\(\displaystyle \frac{dA}{dt}=12(3)(38)=1368\)

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

\(\displaystyle A=6s^2\)

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 2 and a rate of growth of 39:

\(\displaystyle \frac{dA}{dt}=12(2)(39)=936\)