[Math] surface area of a cube related rates problem

derivatives

The volume of a cube is expanding at a rate of $4cm^3/s$. What is the rate the surface area is changing when the area is $24cm^2$.
What I do is:
$$
V=h^3
\\A=6h^2={6V\over{h}}
\\h=\sqrt{A\over6}=2
$$
Then…
$$
{dA\over{dt}}={6{dV\over{dt}}h\over{h^2}}={6\bullet4\bullet2\over{4}}=12cm^2/s
$$
But it says the answer is not 12. Not sure what I did wrong, havent done related rates in a long time. Any help appreciated thanks.

Best Answer

There's a much simpler way to do it. We can solve for $h$ as a function of $V$: $$h=\sqrt[3]{V}$$ Plugging this into the formula for area, we have $$A=6h^2=6V^{2/3}$$ Using the chain rule, $$\frac{\mathrm{d}A}{\mathrm{d}t}=\frac{\mathrm{d}A}{\mathrm{d}V}\cdot\frac{\mathrm{d}V}{\mathrm{d}t}$$ We know $\frac{\mathrm{d}V}{\mathrm{d}t}$. Can you find $\frac{\mathrm{d}A}{\mathrm{d}V}$ and take it from there?

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