<p>The maximum value of the function \(f(x) = 2x^3 - 15x^2 + 36x - 48\) on the set \(A = \{x \mid x^2 + 20 \leq 9x\}\) is:</p>
Step-by-Step Solution
Key Concept: First identify the feasible domain A by solving the inequality x² - 9x + 20 ≤ 0, then find critical points of f(x) within this interval and evaluate f at critical points and endpoints.
<p><strong>Step 1: Find the domain A</strong></p><p>Solve x² + 20 ≤ 9x, which gives x² - 9x + 20 ≤ 0</p><p>Factoring: (x - 4)(x - 5) ≤ 0</p><p>Therefore: A = [4, 5]</p><p><strong>Step 2: Find critical points of f(x)</strong></p><p>f(x) = 2x³ - 15x² + 36x - 48</p><p>f'(x) = 6x² - 30x + 36 = 6(x² - 5x + 6) = 6(x - 2)(x - 3)</p><p>Critical points: x = 2 and x = 3</p><p><strong>Step 3: Check which critical points lie in [4, 5]</strong></p><p>Both x = 2 and x = 3 lie outside A = [4, 5], so we only evaluate at endpoints.</p><p><strong>Step 4: Evaluate f at endpoints</strong></p><p>f(4) = 2(64) - 15(16) + 36(4) - 48 = 128 - 240 + 144 - 48 = -16</p><p>f(5) = 2(125) - 15(25) + 36(5) - 48 = 250 - 375 + 180 - 48 = 7</p><p><strong>Step 5: Determine maximum</strong></p><p>Since f'(x) = 6(x - 2)(x - 3) > 0 for x ∈ [4, 5], f is strictly increasing on [4, 5].</p><p>∴ Maximum value = f(5) = 7</p>
Correct Answer: B