<p>If \(a\), \(b\), and \(c\) are positive and \(9a + 3b + c = 90\), then the maximum value of \((\log a + \log b + \log c)\) is (base of the logarithm is 10) ______.</p>
Step-by-Step Solution
Key Concept: Use AM-GM inequality on the constraint by rewriting 9a + 3b + c = 90 as a weighted sum, then recognize that the product abc is maximized when the weighted terms are equal, which occurs at a specific point that can be found using calculus or AM-GM optimization.
<p><strong>Step 1:</strong> Recognize that log a + log b + log c = log(abc). To maximize this sum, we need to maximize the product abc subject to the constraint 9a + 3b + c = 90.</p><p><strong>Step 2:</strong> Rewrite the constraint as a sum of equal terms by dividing: (9a) + (3b) + (c) = 90. Apply AM-GM inequality:</p><p>$$\frac{9a + 3b + c}{3} \geq \sqrt[3]{9a \cdot 3b \cdot c}$$</p><p><strong>Step 3:</strong> This gives us:</p><p>$$\frac{90}{3} \geq \sqrt[3]{27abc}$$</p><p>$$30 \geq \sqrt[3]{27abc}$$</p><p>$$27000 \geq 27abc$$</p><p>$$1000 \geq abc$$</p><p><strong>Step 4:</strong> Equality in AM-GM holds when 9a = 3b = c. Let 9a = 3b = c = k. Then 3k = 90, so k = 30. This gives a = 10/3, b = 10, c = 30, and abc = 1000.</p><p><strong>Step 5:</strong> The maximum value of log a + log b + log c = log(1000) = log(10³) = 3 log(10) = 3(1) = 3.</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: 3