# Surface area of melting ice block

A cubic block of ice is melting and retains its cubic shape as it melts. Its volume (in $$\rm m^3$$) at time $$t$$ is given by

$$V = 4000-2000 \cdot e^{0.01t}$$

As the ice melts, the bottom face is not exposed to the air. Let the side length of the cube be $$x$$ meters. Let the total surface area of the exposed surfaces of the cube be $$S$$.

At what rate is the exposed surface area decreasing when the cube has lost half of its initial volume?

### My attempt

There are $$5$$ exposed faces of the cube.

At $$t=0$$, $$V=2000$$. So the cube will have $$V=1000$$ when it loses half the volume. So $$x=10$$.

We want:

$$\frac{dS}{dt}= \frac{dS}{dV} \cdot \frac{dV}{dt}$$

First:

$$\frac{dV}{dt}= -0.01 \cdot 2000 \cdot e^{0.01t}$$

$$= -0.01 \cdot (4000-V)$$

$$= \frac{x^3-4000}{100}$$

Now: $$\frac{dS}{dV}= \frac{dS}{dx} \cdot \frac{dx}{dV}$$

By construction:

$$S = 5x^2$$ $$V = x^3$$

Therefore:

$$\frac{dS}{dx} = 10x$$

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

$$\frac{dS}{dV}= \frac{dS}{dx} \cdot \frac{dx}{dV} = \frac{10x}{3x^2} = \frac{10}{3x}$$

Finally:

$$\frac{dS}{dt}= \frac{dS}{dV} \cdot \frac{dV}{dt}$$

$$= \frac{10}{3x} \cdot \frac{x^3-4000}{100}$$

$$= \frac{10}{3 \cdot 10} \cdot \frac{1000-4000}{100}$$

$$= \frac{-3000}{300}$$

$$= -10$$

I'm not sure if this is the solution to the problem OR if it is the best way to attempt this. Are there any more clever methods to solve this one? Thanks in advance.

• Your answer is correct. – Shubham Johri Jun 9 at 9:51