Question

The maximum value and minimum value of the volume of the parallelopiped having coterminous edges \(\hat{\mathrm{i}}+x \hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{j}}+x \hat{\mathrm{k}}\) and \(x \hat{\mathrm{i}}+\hat{\mathrm{k}}\) are respectively

• A. \(\frac{1}{3\sqrt{3}}+1,\frac{-1}{3\sqrt{3}}+1\)
• B. \(\frac{2}{3\sqrt{3}}+1,\frac{-2}{3\sqrt{3}}+1\)
• C. \(\frac{1}{\sqrt{3}}+1,\frac{-1}{\sqrt{3}}+1\)
• D. \(\frac{2}{\sqrt{3}}+1,\frac{-2}{\sqrt{3}}+1\)

Hint:

The volume is the scalar triple product; differentiate it with respect to x to find the extreme values.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Volume \(V=\begin{vmatrix}1&x&1\\0&1&x\\x&0&1\end{vmatrix}\).

Step 2: Expand
\(V=1(1-0)-x(0-x^2)+1(0-x)=x^3-x+1\)

Step 3: Find Critical Points
\(V'=3x^2-1=0\Rightarrow x=\pm\dfrac{1}{\sqrt3}\). Also \(V''=6x\), so \(x=-\dfrac{1}{\sqrt3}\) gives a maximum and \(x=\dfrac{1}{\sqrt3}\) gives a minimum.

Step 4: Calculation
\(V_{\max}=-\dfrac{1}{3\sqrt3}+\dfrac{1}{\sqrt3}+1=1+\dfrac{2}{3\sqrt3}\) and \(V_{\min}=1-\dfrac{2}{3\sqrt3}\).

Final Answer:
The maximum and minimum volumes are 1 plus and minus 2 over 3 root 3. \[ \boxed{\text{(B) }\dfrac{2}{3\sqrt{3}}+1,\ \dfrac{-2}{3\sqrt{3}}+1} \]