Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>A rod of fixed length \(k\) slides along the coordinate axes. If it meets the axes at \(A(a, 0)\) and \(B(0, b)\), then the minimum value of \(\left(a + \dfrac{1}{a}\right)^2 + \left(b + \dfrac{1}{b}\right)^2\) is</p>
<p>(1) 0</p>
<p>(2) 8</p>
<p>(3) \(k^2 - 4 + \dfrac{4}{k^2}\)</p>
<p>(4) \(k^2 + 4 + \dfrac{4}{k^2}\)</p>
Step-by-Step Solution
Key Concept: Use the constraint a² + b² = k² (from the rod's fixed length) combined with the AM-GM inequality applied to the expression (a + 1/a)² + (b + 1/b)². The minimum occurs when symmetry conditions are met: a = b = k/√2.
<p><strong>Step 1:</strong> Set up the constraint. Since the rod of length k slides along the axes meeting at A(a,0) and B(0,b), we have: a² + b² = k²</p><p><strong>Step 2:</strong> Let f(a,b) = (a + 1/a)² + (b + 1/b)². Expand: f = a² + 2 + 1/a² + b² + 2 + 1/b²</p><p><strong>Step 3:</strong> Substitute the constraint a² + b² = k²: f = k² + 4 + 1/a² + 1/b²</p><p><strong>Step 4:</strong> To minimize, we need to minimize 1/a² + 1/b² subject to a² + b² = k². By Cauchy-Schwarz or Lagrange multipliers, the minimum occurs when a = b.</p><p><strong>Step 5:</strong> If a = b, then 2a² = k², so a = b = k/√2</p><p><strong>Step 6:</strong> Substitute back: f = (k/√2 + √2/k)² + (k/√2 + √2/k)² = 2(k/√2 + √2/k)²</p><p><strong>Step 7:</strong> Simplify: k/√2 + √2/k = (k² + 2)/(k√2). Then f = 2 · [(k² + 2)²/(2k²)] = (k² + 2)²/k²</p><p><strong>Step 8:</strong> For k = √2 (or examining the general form), the minimum value is **(k² + 2)²/k² which equals 8 when k = √2**, giving the answer as <strong>8</strong> or an expression depending on given options.</p><p>∴ Answer: D</p>
Correct Answer: D