Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>If <i>a</i> > <i>b</i> > 0 and <i>f</i>(C) = \frac{(<i>a</i><sup>2</sup> − <i>b</i><sup>2</sup>) cos C}{<i>a</i> − <i>b</i> sin C}, then the maximum value of <i>f</i>(C) is</p>
<p>(A) 2√(<i>a</i><sup>2</sup> − <i>b</i><sup>2</sup>)</p>
<p>(B) √(<i>a</i><sup>2</sup> − <i>b</i><sup>2</sup>)</p>
<p>(C) <i>a</i><sup>2</sup> − <i>b</i><sup>2</sup></p>
<p>(D) <i>b</i><sup>2</sup> − <i>a</i><sup>2</sup></p>
Step-by-Step Solution
Key Concept: Rewrite the function as a ratio where maximizing the function is equivalent to minimizing the denominator. Use calculus to find the critical point of the denominator.
<p><strong>Step 1:</strong> We can rewrite <i>f</i>(C) = \frac{<i>a</i><sup>2</sup> − <i>b</i><sup>2</sup>}{<i>h</i>(C)}, where <i>h</i>(C) = <i>a</i> sec C − <i>b</i> tan C.</p><p><strong>Step 2:</strong> <i>f</i>(C) is maximum when <i>h</i>(C) is minimum.</p><p><strong>Step 3:</strong> Taking derivative: <i>h</i>'(C) = sec C (<i>a</i> tan C + <i>b</i> sec C) = 0.</p><p><strong>Step 4:</strong> This gives sin C = \frac{<i>b</i>}{<i>a</i>}.</p><p><strong>Step 5:</strong> Substituting back, the minimum value of <i>h</i>(C) is √(<i>a</i><sup>2</sup> − <i>b</i><sup>2</sup>).</p><p><strong>Step 6:</strong> Therefore, maximum value of <i>f</i>(C) is √(<i>a</i><sup>2</sup> − <i>b</i><sup>2</sup>).</p><p>∴ Answer is (B).</p>
Correct Answer: B