The sides x and y of two squares are related by \(y = 2x - x^{2}\), where \(0 \lt x \lt 1\). The rate of change of the area of the second square with respect to the area of the first square is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2x^{2} - 6x + 4\)
The areas of the first and second squares are
\(A_1 = x^{2}\), so \( \displaystyle \frac{dA_1}{dx} = 2x \)
\(A_2 = y^{2}\), so \( \displaystyle \frac{dA_2}{dx} = 2y\frac{dy}{dx} = 2(2x - x^{2})(2 - 2x) \)
By the chain rule:
\( \displaystyle \frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx} = \frac{2x(2 - x)(2 - 2x)}{2x} \)
\( = (2 - x)(2 - 2x) \)
\( = 4 - 4x - 2x + 2x^{2} \)
\( = 2x^{2} - 6x + 4 \)
Why the other options are incorrect
Option (1) \(2x^{2} + 6x + 4\) — Incorrect. The middle term has the wrong sign: \((2 - x)(2 - 2x)\) gives \(-4x - 2x = -6x\).
Option (3) \(x^{2} - 3x + 2\) — Incorrect. This is half the answer: it takes \(\frac{dA_2}{dx}\) as \(y\frac{dy}{dx}\), losing the factor 2 of \(\frac{d}{dx}y^{2} = 2y\frac{dy}{dx}\).
Option (4) \(2x^{2} - 6x - 4\) — Incorrect. The constant term is \(2 \times 2 = +4\), not −4.