Area Under the Curve
Maximum area of inscribed figure
Grade 12
Question:
<p>A triangle has one vertex at (0, 0) and the other two on the graph of \(y = -2x^2 + 54\) at \((x, y)\) and \((-x, y)\) where \(0 < x < \sqrt{27}\). The value of x so that the corresponding triangle has maximum area is</p>
<p>(A) \(\frac{27}{2}\)</p>
<p>(B) 3</p>
<p>(C) \(2\sqrt{3}\)</p>
<p>(D) None of these</p>
Step-by-Step Solution
Key Concept: Set up area as a function of x and use calculus to find the maximum by setting the derivative to zero.
<p>The triangle has vertices at (0, 0), (x, y) and (-x, y) where $y = -2x^2 + 54$. The base is 2x and height is $y = -2x^2 + 54$. Area $A = x(-2x^2 + 54) = -2x^3 + 54x$. To maximize: $\frac{dA}{dx} = -6x^2 + 54 = 0$ gives $x^2 = 9$, so $x = 3$.</p>
Correct Answer: B