Key Concepts and Formulas

• Optimization using Derivatives: Finding the maximum or minimum of a function by setting its derivative to zero.
• Volume of a Cuboid: $V = lwh$, where $l$ is length, $w$ is width, and $h$ is height.
• Surface Area of a Box (without top): $SA = lw + 2lh + 2wh$
• Product Rule of Differentiation: $(uv)' = u'v + uv'$
• Chain Rule of Differentiation: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Step-by-Step Solution

Step 1: Define Variables and Dimensions

We are given a square tin sheet of side 30 cm. We cut out squares of side $x$ from each corner and fold up the flaps to form an open-top box. We want to maximize the volume of this box and then find its surface area.

• Side of the original square sheet: $L = 30$ cm
• Side of the square cut from each corner: $x$ cm
• Length of the box: $l = 30 - 2x$ cm
• Width of the box: $w = 30 - 2x$ cm
• Height of the box: $h = x$ cm

Step 2: Formulate the Volume Function

The volume of the box is given by $V = lwh$. Substituting the dimensions in terms of $x$, we get:

$V(x) = (30 - 2x)(30 - 2x)x = x(30 - 2x)^2$

To ensure the box is physically possible, the dimensions must be positive. Therefore, $x > 0$ and $30 - 2x > 0$, which implies $x < 15$. Thus, the domain of $x$ is $0 < x < 15$.

Step 3: Find the Critical Points by Differentiating the Volume Function

To find the maximum volume, we need to find the critical points of $V(x)$ by finding where $\frac{dV}{dx} = 0$.

First, differentiate $V(x)$ with respect to $x$ using the product rule:

$\frac{dV}{dx} = \frac{d}{dx} [x(30 - 2x)^2]$

Let $u = x$ and $v = (30 - 2x)^2$. Then $u' = 1$ and $v' = 2(30 - 2x)(-2) = -4(30 - 2x)$ using the chain rule. Applying the product rule gives:

$\frac{dV}{dx} = (1)(30 - 2x)^2 + x[-4(30 - 2x)]$ $\frac{dV}{dx} = (30 - 2x)^2 - 4x(30 - 2x)$ Factor out $(30 - 2x)$: $\frac{dV}{dx} = (30 - 2x)[(30 - 2x) - 4x]$ $\frac{dV}{dx} = (30 - 2x)(30 - 6x)$

Step 4: Solve for x when dV/dx = 0

Set $\frac{dV}{dx} = 0$ to find the critical points:

$(30 - 2x)(30 - 6x) = 0$

This gives two possible values for $x$:

• $30 - 2x = 0 \implies x = 15$
• $30 - 6x = 0 \implies x = 5$

Step 5: Determine the Valid Value of x

Since $x$ must be between 0 and 15, we examine both critical points. $x = 15$ would result in a box with zero length and width, so it is not a valid solution. Therefore, $x = 5$ is the only valid critical point.

Step 6: Calculate the Dimensions of the Box

With $x = 5$, the dimensions of the box are:

• Length: $l = 30 - 2(5) = 20$ cm
• Width: $w = 30 - 2(5) = 20$ cm
• Height: $h = 5$ cm

Step 7: Calculate the Surface Area of the Box (without top)

The surface area of the open-top box is:

$SA = lw + 2lh + 2wh$ $SA = (20)(20) + 2(20)(5) + 2(20)(5)$ $SA = 400 + 200 + 200$ $SA = 800 \text{ cm}^2$

Common Mistakes & Tips

• Domain Restriction: Failing to consider the domain of $x$ can lead to incorrect solutions. Always check that the solution makes physical sense.
• Surface Area Formula: Make sure to use the correct surface area formula for an open-top box.
• Algebraic Errors: Be careful with algebraic manipulations during differentiation and simplification.

Summary

We found the dimensions of the box that maximize its volume by setting the derivative of the volume function equal to zero. The valid critical point was $x=5$ cm, which results in a box with dimensions 20 cm x 20 cm x 5 cm. The surface area of this open-top box is 800 cm$^2$.

The final answer is $\boxed{800}$, which corresponds to option (C).