Brainteaser question/solution which reasonment is correct and why?

Question

I am struggling to find some explanation to this: here is my problem:

"A cube of ice melts without changing shape at uniform rate 4cm$^3$/min. Find the rate of change of the surface area of the cube when the volume is 125cm$^3$"

So the length $L$ of the cube is equal to $5$.

I found two different solutions to this:

my first reasoning is as follow: I compute the volume of the shape after 1 min then I take the the cubic root of the results to find back the length of the cube and I compute the new surface. I subtract the surface before it melts with the new one:

$$ 1) \ V_{new}^{t+1 min} = 125 - 4 = 121\ cm^3$$ $$ 2) \ L_{new} = 121^{\frac13} $$ $$ 3) \ S_{new} = 6\times L_{new}^2$$ $$ 4) \ Rate_S = 150 - S_{new} = 3.21731402243 $$

The second solution: I follow the reasoning with the derivative using the chain rules method.

we know that $\frac{\partial V}{\partial t} = 4\ cm^3/min$ and $V=L^3$ then $$\frac{\partial V}{\partial L} = 3\times L^2$$ we also know that: $$\frac{\partial V}{\partial t} = \frac{\partial V}{\partial L}\frac{\partial L}{\partial t}$$ So $$\frac{\partial L}{\partial t} = \frac{4}{3\times L^2}$$ As for the surface, we know that $S=6\times L^2$ so $$\frac{\partial S}{\partial L} = 12\times L$$ and $$\frac{\partial S}{\partial t} = \frac{\partial S}{\partial L}\frac{\partial L}{\partial t}$$ so $$\frac{\partial S}{\partial t} = 12 \times L \times \frac{4}{3\times L^2} = 3.2$$

Hence here is my problem which one of the two reasoning is wrong? If there is one...

Answer 1

The rate at which the surface area changes is itself constantly changing. Your first method computed the average rate of change over the previous second, but what the question is asking for is the instantaneous rate at the moment that the volume of the cube is 125 ml. This would be the same as the average rate that you computed if the surface area were changing at a constant rate. Observe that if you had used a different time interval, you would’ve gotten a different average rate. Your second method is the correct one to use (although I haven’t checked your arithmetic for either answer).