1. **Stating the problem:** We have a cuboid with volume 360 cm³ and surface area A cm². The dimensions are width = $2x$ cm, height = $y$ cm, and length = $x$ cm. 2. **Expressing $y$ in terms of $x$:** The volume $V$ of a cuboid is given by: $$V = \text{length} \times \text{width} \times \text{height}$$ Substitute the given dimensions: $$360 = x \times 2x \times y = 2x^2 y$$ Solve for $y$: $$y = \frac{360}{2x^2} = \frac{180}{x^2}$$ 3. **Showing that $A = 4x^2 + \frac{1080}{x}$:** The surface area $A$ of a cuboid is: $$A = 2(lw + lh + wh)$$ Substitute $l = x$, $w = 2x$, and $h = y$: $$A = 2(x \times 2x + x \times y + 2x \times y) = 2(2x^2 + xy + 2xy) = 2(2x^2 + 3xy)$$ Simplify: $$A = 4x^2 + 6xy$$ Substitute $y = \frac{180}{x^2}$: $$A = 4x^2 + 6x \times \frac{180}{x^2} = 4x^2 + \frac{1080}{x}$$ 4. **Finding the approximate change in $A$ as $x$ increases from 2 to $2 + p$:** We use the derivative $\frac{dA}{dx}$ to approximate the change: $$A = 4x^2 + \frac{1080}{x}$$ Differentiate: $$\frac{dA}{dx} = 8x - \frac{1080}{x^2}$$ Evaluate at $x=2$: $$\frac{dA}{dx}\bigg|_{x=2} = 8 \times 2 - \frac{1080}{2^2} = 16 - 270 = -254$$ The approximate change in $A$ when $x$ changes by $p$ is: $$\Delta A \approx \frac{dA}{dx} \times p = -254p$$ Since the derivative is negative, the surface area decreases as $x$ increases from 2. **Final answers:** - $y = \frac{180}{x^2}$ - $A = 4x^2 + \frac{1080}{x}$ - Approximate change in $A$ is $-254p$, indicating a decrease when $x$ increases from 2.