Animated Solution for Mathematics - Differentiation: A square piece of tin of side $30\text{ cm}$ is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in ${\text{cm}}^2$) is equal to

Crafting the Perfect Box

Imagine you are standing in a workshop, holding a flat, square piece of tin. It is perfectly smooth, measuring $30\text{ cm}$ on each side.

Your task is simple yet profound: transform this flat sheet into an open-topped box that holds the maximum possible volume. This is not just a geometry problem; it is a dance with optimization, a fundamental concept that underpins everything from engineering efficiency to economic modeling.

Visualizing the Geometry

Before we touch a single equation, let us visualize the process. We start with a sheet of area $A=30\times 30=900{\text{ cm}}^2$.

To create a box, we must cut a square of side $x$ from each of the four corners. When we fold up the resulting flaps, those four squares are discarded.

By removing a square of side $x$ from both ends of each side, the new dimensions of our box become: - Length ($l$) = $30-2x$ - Width ($w$) = $30-2x$ - Height ($h$) = $x$

It is easy to make the mistake of subtracting only $x$, but remember: you are cutting from both ends of the sheet. The geometry demands $30-2x$.

The Mathematical Model

The volume $V$ of a cuboid is defined by the product of its length, width, and height. Substituting our dimensions, we get the volume function:

This is the function we need to maximize. We want to find the value of $x$ that makes $V(x)$ as large as possible. This is where the power of calculus shines.

The Calculus of Maxima

To find the maximum, we look for the critical points by setting the first derivative $V^{\prime }(x)$ to zero. Using the product rule, where $\frac{d}{dx}[u\cdot v]=u^{\prime }v+u{v}^{\prime }$, we differentiate $V(x)=(30-2x{)}^2\cdot x$:

Simplifying this expression is a test of algebraic patience. We factor out $(30-2x)$:

Setting $V^{\prime }(x)=0$ gives us two potential values for $x$: $x=15$ or $x=5$. As we discussed, $x=15$ would result in a length of zero, which is physically impossible. Thus, $x=5$ is our only valid critical point.

The Final Answer

The question asks for the surface area of the box at that specific configuration. The surface area $S$ of our open box is the original area of the sheet minus the four squares we cut away:

Substituting our optimal value $x=5$:

And there you have it! By carefully modeling the geometry and applying the principles of calculus, we have determined that the surface area of the box with maximum volume is $800{\text{ cm}}^2$. You have successfully navigated the constraints of the physical world using the elegance of mathematics.