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A cube of ice is melting at a constant rate . The initial volume of the ice cube is $20cm^3$ and volume after $5$ mins is $15cm^3$. Find the rate at which the length of the cube is decreasing at the time $t=12$mins

My attempt

Let length of the cube = $x$ cm

$\frac{dV}{dt} =( \frac{dV}{dx} )(\frac{dx}{dt})$

$\frac{dV}{dt} = \frac{20-15}{5} = 1cm^3/min$

right now , I understand I need to find $ \frac{dV}{dx} $

I must form an equation of with x

Using ratios ....

$V= (x)^3 $

$ \frac{dV}{dx} = 3x^2 $

Now I'm struggling to find the value of $x$ to substitute into.

I was told

$x^3 = 20 - 12(1) $

But I don't understand the above equation at all . Any help would be appreciated ! Thanks!

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hint

$$V (t)=(x (t))^3$$

$$\frac {dV}{dt}=3x^2 (t)\frac {dx}{dt} $$

$$\implies \frac {dx}{dt}=\frac {dV}{dt}\frac {1}{3 (V (t))^\frac {2}{3}} $$

At $t=12 \; min $,

$V (t)=15 \;cm^3$

$\frac {dV}{dt}=\frac {15-20}{12} \;cm^3/min $

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