Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>From a given solid cone of height \(H\), another inverted cone is carved whose height is \(h\), such that its volume is maximum, then the ratio \(\dfrac{H}{h}\) is equal to:</p>
<p>2</p>
<p>3</p>
<p>4</p>
<p>6</p>
Step-by-Step Solution
Key Concept: Use similar triangles to express the radius of the inverted cone in terms of its height h, then maximize the volume function V(h) by taking its derivative and setting it to zero.
<p><strong>Step 1:</strong> Set up the geometry. Let the original cone have height H and base radius R. The inverted cone has height h (measured from the base of original cone upward into it).</p><p><strong>Step 2:</strong> By similar triangles, the radius of the inverted cone at height h from the base is: r = R(H - h)/H</p><p><strong>Step 3:</strong> The volume of the inverted cone is: V = (1/3)πr²h = (1/3)π[R(H-h)/H]²·h = (πR²/3H²)·(H-h)²·h</p><p><strong>Step 4:</strong> To maximize, differentiate with respect to h: dV/dh = (πR²/3H²)·[(H-h)²·1 + h·2(H-h)·(-1)]</p><p><strong>Step 5:</strong> dV/dh = (πR²/3H²)·[(H-h)² - 2h(H-h)] = (πR²/3H²)·(H-h)[(H-h) - 2h]</p><p><strong>Step 6:</strong> Setting dV/dh = 0: (H-h)(H-3h) = 0. Since h ≠ H, we get: H = 3h</p><p><strong>Step 7:</strong> Therefore: H/h = 3</p><p>∴ Answer: B</p>
Correct Answer: B