Question

A cube of ice melts without changing its shape at the uniform rate of $4c{m}^3/min$. The rate of change of the surface area of the cube, in $c{m}^2/min$, when the volume of the cube is $125c{m}^3$, is

• A. $-4$
• B. $-\frac{16}{5}$
• C. $-\frac{16}{6}$
• D. $-\frac{8}{15}$

Answer 1

The correct option is B $-\frac{16}{5}$

Let the side of the cube be $a$ cms.
Hence, Surface area is $6a^2$ and Volumes is $a^3$
Thus, $V=a^3$
Differentiating with respect to $a$, we get
$\frac{dV}{dt}=3a^2.\frac{da}{dt}\to$(i)
Now volume at that instant is $125c{m}^3$
Or $a^3=125c{m}^3$
Or $a=5cm$
Substituting in (i), we get
$\frac{dV}{dt}=3\left(25\right).\frac{da}{dt}$
$4=75.\frac{da}{dt}$
Or $\frac{da}{dt}=\frac{4}{75}cm/min$
Now
$S=6a^2$
$\frac{dS}{dt}=12a\cdot \frac{da}{dt}=60\cdot \left(\frac{4}{75}\right)=\frac{16}{5}$
Since it is decreasing, we put a negative sign,
$\frac{dS}{dt}=-\frac{16}{5}c{m}^2/min$