Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>The differential equation \(y^3 y'' = -1\) with \(y'y'' = \frac{-y'}{y^3}\), is solved with conditions \(x=1,\ y=1\) and \(y'(1)=0\). The function \(f(x) = y = \sqrt{2x - x^2}\) is defined for \(y \geq 0,\ x \in D_f\). What is the maximum value of \(f(x)\)?</p>
<p>\(1\)</p>
<p>\(2\)</p>
<p>\(\sqrt{2}\)</p>
<p>\(\frac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the given differential equation y³y'' = -1 describes a curve, and solving with initial conditions y(1)=1, y'(1)=0 yields the explicit function f(x) = √(2x - x²). The maximum occurs where the derivative equals zero or at domain boundaries.
<p><strong>Step 1:</strong> Rewrite f(x) = √(2x - x²) = √(1 - (x-1)²) by completing the square.</p><p><strong>Step 2:</strong> Find the domain: 2x - x² ≥ 0 → x(2-x) ≥ 0 → D_f = [0, 2]</p><p><strong>Step 3:</strong> Recognize that f(x) = √(1 - (x-1)²) represents the upper semicircle with center (1,0) and radius 1.</p><p><strong>Step 4:</strong> To find the maximum, take the derivative: f'(x) = (2 - 2x)/(2√(2x - x²)) = (1-x)/√(2x - x²)</p><p><strong>Step 5:</strong> Set f'(x) = 0: This gives 1 - x = 0 → x = 1</p><p><strong>Step 6:</strong> Evaluate: f(1) = √(2(1) - 1²) = √(2 - 1) = √1 = 1</p><p><strong>Step 7:</strong> Check boundaries: f(0) = 0, f(2) = 0. The maximum occurs at x = 1.</p><p>∴ Answer: <strong>Maximum value of f(x) = 1</strong></p>
Correct Answer: A