The surface area of a cube is 150 sq. cm. If it is increased by 0.025 sq. cm, then the approximate increase in its volume (in c.c.) is:
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When asked for an approximate change, use differentials (\(dV \approx V'(x)dx\)) rather than calculating the difference between two exact values. This method is much faster and accurate for small changes.
Step 1: Write the cube formulas.
For side $x$, surface area $S=6x^2$ and volume $V=x^3$. We want the small change $dV$ for a small change $dS=0.025$. Step 2: Find the current side.
$6x^2=150$ gives $x^2=25$, so $x=5$ cm. Step 3: Differentiate both formulas.
$\dfrac{dS}{dx}=12x$ and $\dfrac{dV}{dx}=3x^2$. Step 4: Relate $dV$ to $dS$.
Dividing, $\dfrac{dV}{dS}=\dfrac{3x^2}{12x}=\dfrac{x}{4}$. Step 5: Plug in numbers.
\[ dV=\frac{x}{4}\,dS=\frac{5}{4}(0.025)=1.25\times0.025. \] Step 6: Compute.
\[ dV=0.03125\ \text{c.c.} \]
\[ \boxed{0.03125} \]