Question

1. multiple choice if the volume of a cube is increasing at 24 in³/min and each edge of the cube is increasing at 2 in./min, what is the length of each edge of the cube? (a) 2 in. (b) 2√2 in. (c) ∛12 in (d) 4 in. (e) 8 in. 39. multiple choice if the volume of a cube is increasing at 24 in³/min and the surface area of the cube is increasing at 12 in²/ min, what is the length of each edge of the cube? (a) 2 in. (b) 2√2 in. (c) ∛12 in. (d) 4 in. (e) 8 in.

Explanation:

Step1: Recall volume formula for cube

Let the edge - length of the cube be $x$. The volume of a cube $V = x^{3}$. Differentiate with respect to time $t$ using the chain - rule: $\frac{dV}{dt}=3x^{2}\frac{dx}{dt}$.

Step2: Solve for $x$ in the first problem

We know that $\frac{dV}{dt}=24$ in³/min and $\frac{dx}{dt}=2$ in/min. Substitute these values into $\frac{dV}{dt}=3x^{2}\frac{dx}{dt}$:
\[

$$\begin{align*} 24&=3x^{2}\times2\\ 24& = 6x^{2}\\ x^{2}&=4\\ x& = 2\text{ in} \end{align*}$$

\]

Step3: Recall surface - area formula for cube

The surface - area of a cube $S = 6x^{2}$. Differentiate with respect to time $t$ using the chain - rule: $\frac{dS}{dt}=12x\frac{dx}{dt}$.

Step4: Recall volume formula and differentiate

We still have $V = x^{3}$, so $\frac{dV}{dt}=3x^{2}\frac{dx}{dt}$, and we know that $\frac{dV}{dt}=24$ in³/min and $\frac{dS}{dt}=12$ in²/min.
From $\frac{dS}{dt}=12x\frac{dx}{dt}=12$, we get $\frac{dx}{dt}=\frac{1}{x}$.
Substitute $\frac{dx}{dt}=\frac{1}{x}$ into $\frac{dV}{dt}=3x^{2}\frac{dx}{dt}$:
\[

$$\begin{align*} 24&=3x^{2}\times\frac{1}{x}\\ 24&=3x\\ x& = 8\text{ in} \end{align*}$$

\]

Answer:

1. A. 2 in.
2. E. 8 in.