surface area of a cube related rates problem

Question

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.

Answer 1

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?

Answer 2

An alternative way:

$$ V = h^3 ⇔ \frac{dV}{dh} = 3h^2 $$

$$ A = 6h^2 ⇔ \frac{dA}{dh} = 12h $$

$$ \frac{dV}{dt} = 4 $$

Therefore $$ \frac{dA}{dt} = \frac{dA}{dh} * \frac{dh}{dV} * \frac{dV}{dt} $$

$$ \frac{dA}{dt} = 12h * \frac{1}{3h^2} * 4 $$

$$ \frac{dA}{dt} = \frac{16}{h} $$

When Area = $24$

$$ A = 6h^2 $$

$$ 24 = 6h^2 $$

$$ h = 2 $$

$$\frac{dA}{dt}_{h=2} = \frac{16}{h} $$

$$\frac{dA}{dt} = 8 cm^{2}/s $$