Chapter 4.6, Problem 38E

To determine

To find: The edge of a cube when rate in increasing of volume and edge is given.

Answer to Problem 38E

The option (A) is correct.

Explanation of Solution

Given data: The volume of the cube is increasing at $24\text{in}{\text{.}}^{\text{3}}/\min$ and each edge increasing at 2 in. /min. Then the length of the cube is,

  $\left(\text{A}\right)2\text{in}\text{.}\left(\text{B}\right)2\sqrt{2}\text{in}\text{.}\left(\text{C}\right)\sqrt[3]{12}\text{in}\text{.}\left(\text{D}\right)8\text{in}\text{.}$

Method/Formula used:

The volume V of a cube with edge is given by $V=a^3$ .

Calculation:

The volume of cube is increasing at $24\text{in}{\text{.}}^{\text{3}}/\min$ and each edge increasing at 2 in. /min. i.e.,

  $\left.\begin{array}{c}\frac{dV}{dt}=24\text{in}{\text{.}}^{\text{3}}\text{/min}\text{.}\\ \frac{da}{dt}=\text{2}\text{in}\text{./min}\text{.}\end{array}\right\}$ … (1)

The volume of the cube is,

  $V=a^3$

Differentiating with respect to ‘ $t$ ’,

  $\left.\begin{array}{c}\frac{dV}{dt}=\frac{d}{dt}\left(a^3\right)\\ =3a^2\frac{da}{dt}\end{array}\right\}$ … (2)

Substitute $24\text{in}{\text{.}}^{\text{3}}\text{/min}\text{.}$ for $\frac{dV}{dt}$ and $\text{2}\text{in}\text{./min}\text{.}$ for $\frac{da}{dt}$ in result (2),

  $\begin{array}{c}24\text{in}{\text{.}}^{\text{3}}\text{/min}\text{.}=3a^2\left(2\text{in}\text{./min}\text{.}\right)\\ a^2=\frac{24\text{in}{\text{.}}^{\text{3}}\text{/min}\text{.}}{2\cdot \left(2\text{in}\text{./min}\right)}\\ a^2=4\text{in}{\text{.}}^{\text{2}}\\ a=2\text{in}\text{.}\end{array}$

Thus, the edge of the cube is $2\text{in}\text{.}$ .

Therefore, the option (A) is correct.