Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length \(x\). The maximum area enclosed by the park is</p>
<p>\(\dfrac{3}{2}x^2\)</p>
<p>\(\sqrt{\dfrac{x^3}{8}}\)</p>
<p>\(\dfrac{1}{2}x^2\)</p>
<p>\(\pi x^2\)</p>
Step-by-Step Solution
Key Concept: For a triangular park with two equal sides of length x and a fixed perimeter constraint from the river, the area is maximized when the triangle is isosceles right-angled. Use the relationship between the two equal sides and the base to express area as a function of one variable, then apply calculus.
<p><strong>Step 1:</strong> Set up the problem. Two sides of equal length x meet at angle θ. The third side (along river) is unrestricted.</p><p><strong>Step 2:</strong> Express area as a function of θ. For two sides of length x with included angle θ:<br/>A(θ) = (1/2)·x·x·sin(θ) = (x²/2)sin(θ)</p><p><strong>Step 3:</strong> Find the maximum by taking derivative with respect to θ:<br/>dA/dθ = (x²/2)cos(θ)</p><p><strong>Step 4:</strong> Set dA/dθ = 0:<br/>cos(θ) = 0 ⟹ θ = 90°</p><p><strong>Step 5:</strong> Verify this is a maximum using second derivative:<br/>d²A/dθ² = -(x²/2)sin(θ) < 0 at θ = 90°, confirming maximum.</p><p><strong>Step 6:</strong> Calculate maximum area:<br/>A_max = (x²/2)sin(90°) = (x²/2)·1 = <strong>x²/2</strong></p><p>∴ Answer: C</p>
Correct Answer: C