<p>The sum of two positive quantities is equal to 2<em>n</em>. Find the probability that their product is not less than 3/4 times their greatest product.</p>
Step-by-Step Solution
Key Concept: Let the two positive quantities be x and (2n-x). Their product P(x) = x(2n-x) is maximized when x = n, giving maximum product n². We need P(x) ≥ (3/4)n², which translates to finding the favorable interval length divided by the total interval length [0, 2n].
<p><strong>Step 1:</strong> Let the two positive quantities be x and (2n-x), where 0 < x < 2n.</p><p><strong>Step 2:</strong> The product P(x) = x(2n-x) = 2nx - x². This is maximized at x = n, giving maximum product P_max = n².</p><p><strong>Step 3:</strong> We need P(x) ≥ (3/4)n², so: x(2n-x) ≥ (3/4)n²</p><p><strong>Step 4:</strong> Rearranging: 2nx - x² ≥ (3/4)n² → x² - 2nx + (3/4)n² ≤ 0</p><p><strong>Step 5:</strong> Using the quadratic formula: x = (2n ± √(4n² - 3n²))/2 = (2n ± n)/2</p><p>So x = (3n/2) or x = (n/2)</p><p><strong>Step 6:</strong> The inequality x² - 2nx + (3/4)n² ≤ 0 is satisfied when n/2 ≤ x ≤ 3n/2.</p><p><strong>Step 7:</strong> The favorable interval length is 3n/2 - n/2 = n. The total interval length is 2n.</p><p><strong>Step 8:</strong> Probability = n/(2n) = 1/2</p><p>∴ Answer: C (1/2)</p>
Correct Answer: C