Applications of Derivatives
Maxima and Minima
Grade 12

Question:

<p>The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is:</p>
<p>\(\sqrt{6}\)</p>
<p>\(\dfrac{2}{3}\sqrt{3}\)</p>
<p>\(2\sqrt{3}\)</p>
<p>\(\sqrt{3}\)</p>

Step-by-Step Solution

Key Concept: For a cylinder inscribed in a sphere, express the radius of the cylinder in terms of its height using the sphere equation, then maximize volume by taking derivatives and setting them to zero.
<p><strong>Step 1:</strong> Set up the constraint. For a cylinder of radius <em>r</em> and height <em>h</em> inscribed in a sphere of radius 3, the cylinder's circular edges touch the sphere. The center of the sphere is at the cylinder's axis, so: r² + (h/2)² = 9</p><p><strong>Step 2:</strong> Express volume in terms of h alone. From the constraint: r² = 9 - h²/4</p><p>Volume V = πr²h = π(9 - h²/4)h = π(9h - h³/4)</p><p><strong>Step 3:</strong> Maximize by taking derivative. dV/dh = π(9 - 3h²/4) = 0</p><p>This gives: 9 = 3h²/4 → h² = 12 → h = 2√3</p><p><strong>Step 4:</strong> Verify this is a maximum. d²V/dh² = π(-3h/2) < 0 at h = 2√3 ✓</p><p>∴ Answer: <strong>C</strong> (h = 2√3)</p>
Correct Answer: C

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