Area Under the Curve
Maxima/Minima of Enclosed Areas
Grade 12
Question:
<p>The area (in sq units) of the largest rectangle ABCD whose vertices A and B lie on the X-axis and vertices C and D lie on the parabola, \(y = x^2 - 1\) below the X-axis, is</p>
<p>(a) \(\frac{4}{3\sqrt{3}}\)</p>
<p>(b) \(\frac{2}{3\sqrt{3}}\)</p>
<p>(c) \(\frac{1}{3\sqrt{3}}\)</p>
<p>(d) \(\frac{4}{3}\)</p>
Step-by-Step Solution
Key Concept: Find the dimensions of the rectangle using symmetry, then use calculus to maximize the area function.
<p><strong>Solution:</strong></p><p>Equation of given parabola: $y = x^2 - 1$</p><p>By symmetry, let the coordinates be: $A(-a, 0)$, $B(a, 0)$, $C(a, a^2 - 1)$, and $D(-a, a^2 - 1)$</p><p>Area of rectangle: $P(a) = 2a(a^2 - 1)$</p><p>For maxima: $P'(a) = 0$</p><p>$\frac{d}{da}[2a(a^2 - 1)] = 2(a^2 - 1) + 4a^2 = 0$</p><p>$6a^2 - 2 = 0$</p><p>$a = \frac{1}{\sqrt{3}}$</p><p>Maximum area = $2 \cdot \frac{1}{\sqrt{3}} \cdot \left(\frac{1}{3} - 1\right) = \frac{2}{\sqrt{3}} \cdot \left(-\frac{2}{3}\right) = \frac{4}{3\sqrt{3}}$</p><p>∴ Answer is (a).</p>
Correct Answer: A