# The volume of a cube is increasing at the rate of 8cm^{3} / s. How fast is the surface area increasing when the length of its edge is 12 cm?

**Solution:**

In maths, derivatives have wide usage. They are used in many situations like finding maxima or minima of a function, finding the slope of the curve, and even inflection point

Let the side length, volume and surface area respectively be equal to x , V and S.

Hence, V = x^{3} and S = 6x^{2}

We have,

dV/dt = 8cm^{3}/ s

Therefore,

dV/dt = d/dt (x^{3})

8 = d/dt (x^{3}) dx/dt

8 = 3x^{2} dx/dt

dx/dt = 8/3x^{2}-------(1)

Now,

dS/dt = d/dt (6x^{2})

= d/dx (6x^{2}) dx/dt

= 12x dx/dt

= 12x (8/3x^{2}) [from (1)]

So, when x = 12 cm

Then,

dS/dt = 32/12 cm^{2}/s

= 8/3 cm^{2}/s

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.1 Question 2

## The volume of a cube is increasing at the rate of 8cm^{3} / s. How fast is the surface area increasing when the length of its edge is 12 cm ?

**Summary:**

Given that The volume of a cube is increasing at the rate of 8cm^{3} / s. Hence, the rate at which surface area increasing when the length of its edge is 12 cm is 8/3 cm^{2}/s