<p><strong>715.</strong> Let \( x,\, y,\, z \) and \( t \) be real numbers such that \( (x,\, y) \) lies on a circle having radius 3; \( (z,\, t) \) lies on a circle having radius 2 and \( xt - yz = 6 \). Find the greatest value of \( P = xz \).<br>[Note: Both circles have centre at origin.]</p>
Step-by-Step Solution
Key Concept: Use the constraint xt - yz = 6 to express one variable in terms of others, then parameterize both circles as x = 3cosα, y = 3sinα, z = 2cosβ, t = 2sinβ to find the maximum of xz = 6cosαcosβ subject to the cross-product constraint.
<p><strong>Step 1:</strong> Parameterize the circles. Since (x,y) lies on circle of radius 3 and (z,t) on circle of radius 2 (both centered at origin):</p><p>x = 3cosα, y = 3sinα</p><p>z = 2cosβ, t = 2sinβ</p><p><strong>Step 2:</strong> Apply the constraint xt - yz = 6:</p><p>3cosα · 2sinβ - 3sinα · 2cosβ = 6</p><p>6(cosαsinβ - sinαcosβ) = 6</p><p>sin(β - α) = 1</p><p>Therefore: β - α = π/2, so β = α + π/2</p><p><strong>Step 3:</strong> Express P = xz in terms of α alone:</p><p>P = 3cosα · 2cos(α + π/2)</p><p>P = 6cosα · (-sinα)</p><p>P = -6cosαsinα = -3sin(2α)</p><p><strong>Step 4:</strong> Find the maximum value:</p><p>Since -3sin(2α) has range [-3, 3], the maximum occurs when sin(2α) = -1:</p><p>P<sub>max</sub> = -3(-1) = 3</p><p>This occurs when 2α = -π/2, i.e., α = -π/4</p><p><strong>Verification:</strong> At α = -π/4: x = 3/√2, y = -3/√2, z = -√2, t = 1, and xz = -3·(-1/√2)·√2 = 3 ✓</p><p>∴ <strong>Answer: 3</strong></p>
Correct Answer: 3