Question

The volume of a cube is increasing at a rate of 181818 cubic meters per hour. At a certain instant, the volume is 888 cubic meters. What is the rate of change of the surface area of the cube at that instant (in square meters per hour)?

Answer 1

Answer:

Rate of change of the surface area is 36 m² per hour.

Step-by-step explanation:

Let x be the side of the cube,

So, the volume of the cube,

$V=x^3$

$\implies x = V^\frac{1}{3}----(1)$

Since, the surface area of the cube,

$A=6x^2$

$\implies A=6V^\frac{2}{3}$ ( From equation (1) )

Differentiating with respect to t ( time ),

$\frac{dA}{dt}=\frac{2}{3}6 V^{-\frac{1}{3}}\frac{dV}{dt}$

We have,

V = 8 meters, $\frac{dV}{dt}$ = 18 m³ / h,

$\frac{dA}{dt}=\frac{12}{3}\times (8)^{-\frac{1}{3}}\times 18$

$=\frac{72}{8^\frac{1}{3}}$

$=\frac{72}{2}$

$=36\text{ square meters per hour}$

Answer 2

Answer:

on khan it is 18

Step-by-step explanation: