Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p><strong>51.</strong> A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = <em>x</em> units and a circle of radius = <em>r</em> units. If the sum of the areas of the square and the circle so formed is minimum, then:</p>
<p>(A) \(2x = r\)</p>
<p>(B) \(2x = (\pi + 4)r\)</p>
<p>(C) \((4 - \pi)x = \pi r\)</p>
<p>(D) \(x = 2r\)</p>
Step-by-Step Solution
Key Concept: Express total area as a function of one variable using the constraint that total wire length = 2, then minimize using calculus. The wire is divided into two parts: one forms square perimeter (4x) and other forms circle circumference (2πr).
<p><strong>Step 1:</strong> Set up the constraint equation. Wire length = 2 units</p><p>For square: perimeter = 4x, so 4x is wire used for square</p><p>For circle: circumference = 2πr, so 2πr is wire used for circle</p><p>Constraint: 4x + 2πr = 2 ... (1)</p><p><strong>Step 2:</strong> Express the total area A</p><p>A = x² + πr²</p><p><strong>Step 3:</strong> Express one variable in terms of the other using constraint (1)</p><p>From (1): 4x + 2πr = 2</p><p>r = (2 - 4x)/(2π) = (1 - 2x)/π</p><p><strong>Step 4:</strong> Substitute into area formula</p><p>A(x) = x² + π[(1 - 2x)/π]²</p><p>A(x) = x² + (1 - 2x)²/π</p><p>A(x) = x² + (1 - 4x + 4x²)/π</p><p><strong>Step 5:</strong> Differentiate and find critical point</p><p>dA/dx = 2x + (−4 + 8x)/π = 0</p><p>2x + (8x − 4)/π = 0</p><p>2πx + 8x − 4 = 0</p><p>x(2π + 8) = 4</p><p>x = 4/(2π + 8) = 2/(π + 4)</p><p><strong>Step 6:</strong> Find corresponding r</p><p>From r = (1 − 2x)/π:</p><p>r = [1 − 4/(π + 4)]/π = [(π + 4 − 4)/(π + 4)]/π = 1/(π + 4)</p><p><strong>Step 7:</strong> Verify relationship</p><p>x/r = [2/(π + 4)]/[1/(π + 4)] = 2</p><p>Therefore: <strong>x = 2r</strong></p><p>∴ Answer: D</p>
Correct Answer: D