Applications of Derivatives
Minimum perimeter / optimization
Grade 12
Question:
<p>Consider a triangle ABC with vertices A, B, C so that A = (a, b), B lies on the x-axis, and C lies on the line y = x. Further let D = (a, -b) be the reflection of A in the x-axis, and let E = (b, a) be the reflection of A in the line y = x. Then the minimum perimeter of triangle ABC is</p>
<p>(a) \(\sqrt{2a^2 + 2b^2}\)</p>
<p>(b) \(\sqrt{2(a-b)^2 + (a+b)^2}\)</p>
<p>(c) \(\sqrt{(a-b)^2 + (a+b)^2}\)</p>
<p>(d) \(2\sqrt{a^2 + b^2}\)</p>
Step-by-Step Solution
Key Concept: The perimeter of triangle ABC equals AB + BC + CA. Since B lies on the x-axis and C lies on y = x, use reflections D and E to convert this into a path length problem: perimeter = AD + DE + EA (by reflection properties), which is minimized when D, E, and the path points are collinear.
<p><strong>Step 1:</strong> Identify key coordinates: A = (a, b), B on x-axis so B = (t, 0), C on line y = x so C = (s, s).</p><p><strong>Step 2:</strong> Use reflection principle. The reflection of A in x-axis is D = (a, -b), and reflection in y = x is E = (b, a).</p><p><strong>Step 3:</strong> By reflection properties: AB = DB (B on x-axis is the mirror), and AC = EC (C on y = x is the mirror). Therefore, Perimeter = AB + BC + CA = DB + BC + CE.</p><p><strong>Step 4:</strong> The perimeter is minimized when D, B, C, E are collinear (straight line gives minimum distance). The minimum perimeter equals the distance DE.</p><p><strong>Step 5:</strong> Calculate DE: DE = √[(b - a)² + (a - (-b))²] = √[(b - a)² + (a + b)²] = √[b² - 2ab + a² + a² + 2ab + b²] = √[2a² + 2b²] = √2 · √(a² + b²).</p><p><strong>Step 6:</strong> The minimum perimeter of triangle ABC is <strong>√2(a² + b²)^(1/2)</strong> or equivalently <strong>√(2(a² + b²))</strong>.</p><p>∴ Answer: AB</p>
Correct Answer: AB