Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>Twenty metres of wire is available for fencing off a flower-bed in the form of a circular sector. Then, the maximum area (in sq. m) of the flower-bed, is</p>
<p>10</p>
<p>25</p>
<p>30</p>
<p>12.5</p>
Step-by-Step Solution
Key Concept: For a circular sector with fixed perimeter (20 m), the area A = (1/2)r²θ is maximized when we express θ in terms of r using the constraint 2r + rθ = 20, then optimize using calculus.
<p><strong>Step 1:</strong> Set up the constraint. For a circular sector with radius r and angle θ (in radians), the perimeter is the two radii plus arc length:</p><p>2r + rθ = 20</p><p>Therefore: θ = (20 - 2r)/r</p><p><strong>Step 2:</strong> Express area in terms of r only. Area of sector: A = (1/2)r²θ</p><p>A = (1/2)r² · (20 - 2r)/r = (1/2)r(20 - 2r) = 10r - r²</p><p><strong>Step 3:</strong> Maximize by taking derivative with respect to r:</p><p>dA/dr = 10 - 2r = 0</p><p>r = 5 m</p><p><strong>Step 4:</strong> Verify this is maximum (d²A/dr² = -2 < 0 ✓) and calculate maximum area:</p><p>A<sub>max</sub> = 10(5) - 5² = 50 - 25 = <strong>25 sq. m</strong></p><p><strong>Step 5:</strong> Check constraint: θ = (20 - 10)/5 = 2 radians, and perimeter = 2(5) + 5(2) = 20 ✓</p><p>∴ Answer: B (25 sq. m)</p>
Correct Answer: B