Question

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm, the ice-cream melts at the rate of 81 cm³/min and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4\pi}$ cm/min. The surface area (in cm²) of the chocolate ball (without the ice-cream layer) is:

Answer 1

256π cm²

Answer 2

Let the chocolate ball have radius $R$ and let the ice‐cream layer thickness be $t$. Then, the volume of the ice‐cream layer is:

$$V = \frac{4}{3}\pi\left[(R+t)^3 - R^3\right].$$

Differentiating with respect to time $t$ (noting that $R$ is constant) gives:

$$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3(R+t)^2 \frac{dt}{dt} = 4\pi (R+t)^2 \frac{dt}{dt}.$$

At the moment when $t = 1$ cm, it is given that:

$$\frac{dV}{dt} = -81 \text{ cm}^3/\text{min} \quad \text{and} \quad \frac{dt}{dt} = -\frac{1}{4\pi} \text{ cm/min}.$$

Substitute these into the differentiated equation:

$$-81 = 4\pi (R+1)^2 \left(-\frac{1}{4\pi}\right).$$

Simplify:

$$-81 = -(R+1)^2 \quad \Rightarrow \quad (R+1)^2 = 81.$$

Taking the positive square root ($R$ and $t$ are positive),

$$R+1 = 9 \quad \Rightarrow \quad R = 8 \text{ cm}.$$

The surface area of the chocolate ball (without the layer) is:

$$\text{Surface Area} = 4\pi R^2 = 4\pi (8)^2 = 256\pi \text{ cm}^2.$$