Problem 90 - Calculus for JEE ADVANCED

Question

The dimensions of the box after cutting equal squares of side \(x\) on the comer will be \(21-2 x, 16-2 x\), and height \(x\). $$ \begin{aligned} V &=x(21-2 x)(16-2 x) \\ &=x\left(336-74 x+4 x^{2}\right)=4 x^{3}+336 x-74 x^{2} \end{aligned} $$

Step-by-Step Solution

Answer

The box's volume is \(4x^3 - 74x^2 + 336x\).

Step 1: Identify Box Dimensions

The problem states that we cut equal squares of side length \(x\) from each corner of a rectangular piece to form a box. The resulting dimensions are: length \((21 - 2x)\), width \((16 - 2x)\), and height \(x\).

Step 2: Understand Volume Calculation

The volume \(V\) of the box is calculated by multiplying the dimensions: height times length times width. Thus, \(V = x(21-2x)(16-2x)\).

Step 3: Expand the Volume Expression

Expand the volume expression by multiplying \((21-2x)(16-2x)\) first: \(21 \times 16 = 336\), \( -2x \times 21 = -42x\), \( -2x \times 16 = -32x\), and \( (-2x) \times (-2x) = 4x^2\). Combining like terms gives \(336 - 74x + 4x^2\).

Step 4: Multiply by Height

Now multiply the expression \((336 - 74x + 4x^2)\) by \(x\) to find the volume: \(x \times 336 = 336x\), \(x \times -74x = -74x^2\), and \(x \times 4x^2 = 4x^3\).

Step 5: Combine like terms for Volume

Combine all terms from the multiplication: \(4x^3 - 74x^2 + 336x\). This results in the final expression for the volume of the box.

Key Concepts

Volume Calculation

Calculating the volume of a box is an essential skill in calculus and everyday math. To find the volume, we simply multiply the length, width, and height of the box. Each of these dimensions is crucial to determining how much space the box encloses.

In this exercise, the volume formula used is:

• Volume, \( V = \, \text{height} \times \text{length} \times \text{width} \)

If dimensions change, the volume updates accordingly. This is why understanding how to adjust dimensions, based on given conditions like cutting squares from a corner, is important. Doing so affects the volume, just as it does in our task here. The task simplifies to finding the new dimensions after modification and performing the multiplication.

Box Dimensions

The box's dimensions in this problem originate from modifying a rectangular piece. Imagine cutting a square from each corner; you can visualize this leading to new edges folding up to form the box's sides.

Initially, the full piece has certain measurements, but after creating sides, these change.

• The resultant length: \( 21 - 2x \)
• The resultant width: \( 16 - 2x \)
• The height will be: \( x \)

These measurements are essential for further calculations and understanding what our starting parameters truly are. It is the critical first step before moving on to calculate the box's volume.

Expansion of Expressions

The expansion of expressions often arises in algebra when finding out comprehensive expressions from multiplied terms. In our exercise, we need to expand \((21-2x)(16-2x)\), which emerges from our box dimensions.

Here's how we do it:

• Multiply the terms: start with distributing the first term (21) to the second binomial \( (16 - 2x) \)
• Do the same for \(-2x \) with \( (16 - 2x) \)

This yields: - \( 21 \times 16 = 336 \)- \( 21 \times -2x = -42x \)- \( -2x \times 16 = -32x \)- \( (-2x) \times (-2x) = 4x^2 \)
Combine these into \( 336 - 74x + 4x^2 \) to proceed with calculating volume.

Algebraic Multiplication Techniques

Algebraic multiplication involves various techniques, in our context it's crucial for volume equations. This involves distributed property application, ensuring each term is considered and correctly multiplied.

Once we have the expanded expression's components, multiply each term by the height, \( x \), of the box. Begin with:

• Multiplying \( x \times 336 = 336x \)
• Then: \( x \times -74x = -74x^2 \)
• And finally: \( x \times 4x^2 = 4x^3 \)

Each step requires precision, maintaining the order of terms, ensuring like terms such as those involving \( x \) or \( x^2 \) are systematically simplified.
This results in the final, combined equation for volume: \( 4x^3 - 74x^2 + 336x \).