2. The volume of a cube is increasing at the rate of \( 12 \) in\(^3\)/min. How fast is the surface area increasing, in square inches per minute, when the length of an edge is \( 20 \) in?






Answer is: option4

\( \frac{12}{5} \)

Solution:

\( V = s^3 \)

\(\frac{dV}{dt} = 3s^2 \frac{ds}{dt} \)

\( 12 = 3 \times (20)^2 \times \frac{ds}{dt} \)

\(\frac{ds}{dt} = \frac{12}{3 \times 400} = 0.01 \)

\( A = 6s^2 \)

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

\(\frac{dA}{dt} = 12 \times 20 \times 0.01 \)

\(\frac{12}{5} \)

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