Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>A wire of length 2 units is cut into two parts which are bent, respectively, to form a square of side \(x\) units and a circle of radius \(r\) units. If the sum of the areas of the square and the circle so formed is minimum, then</p>
<p>\(2x = r\)</p>
<p>\(2x = (\pi + 4)r\)</p>
<p>\((4 - \pi)x = \pi r\)</p>
<p>\(x = 2r\)</p>
Step-by-Step Solution
Key Concept: Express the constraint (total wire length = 2) to eliminate one variable, then minimize the combined area function using calculus. The critical point where dA/dx = 0 gives the minimum.
<p><strong>Step 1:</strong> Set up the constraint. Let wire length for square = 4x, and for circle = 2πr. Then: 4x + 2πr = 2, so x + (πr/2) = 1/2 ... (i)</p><p><strong>Step 2:</strong> Express area function. Total area A = x² + πr². From (i): x = 1/2 - πr/2. Substitute:</p><p>A(r) = (1/2 - πr/2)² + πr² = 1/4 - (π/2)r + (π²/4)r² + πr²</p><p><strong>Step 3:</strong> Minimize by taking derivative: dA/dr = -(π/2) + (π²/2)r + 2πr = 0</p><p>dA/dr = -(π/2) + r(π²/2 + 2π) = 0</p><p>r(π + 4)π/2 = π/2</p><p>r = 1/(π + 4)</p><p><strong>Step 4:</strong> From constraint: x = 1/2 - π/2 · 1/(π + 4) = 1/2 · (π + 4 - π)/(π + 4) = 2/(π + 4)</p><p><strong>Verify:</strong> d²A/dr² > 0 confirms minimum. The ratio x/r = 2/(π + 4) ÷ 1/(π + 4) = 2, so <strong>x = 2r</strong></p><p>∴ Answer: D</p>
Correct Answer: D