Problem 14 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A box with no top is to be made from a \(22 \mathrm{~cm}\) by \(28 \mathrm{~cm}\) piece of card board by cutting squares of equal size from each corner and folding up the 'tabs'. What size of squares should be cut from each corner to make the box of largest volume?

Step-by-Step Solution

Answer

Cut squares of approximately 3.41 cm from each corner for maximum volume.

Step 1: Understand the Problem

We need to create a box with no top from a piece of cardboard that measures \( 22 \text{ cm} \) by \( 28 \text{ cm} \). By cutting squares of side \( x \) cm from each corner and folding the sides up, the goal is to determine \( x \) such that the volume of the box is maximized.

Step 2: Express Volume in Terms of x

Once the squares are cut from each corner, the dimensions of the box will be \( (22-2x) \) cm by \( (28-2x) \) cm with a height of \( x \) cm. The volume \( V \) of the box can be expressed as:\[ V = x(22-2x)(28-2x) = x(616 - 100x + 4x^2) \]

Step 3: Simplify the Volume Equation

Expand the equation:\[ V = 4x^3 - 100x^2 + 616x \]This is the cubic function in terms of \( x \) that represents the volume of the box.

Step 4: Determine Critical Points

To find the maximum volume, take the derivative of the volume and set it equal to zero:\[ V' = 12x^2 - 200x + 616 \]Set \( V' = 0 \) to find critical points:\[ 12x^2 - 200x + 616 = 0 \]

Step 5: Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 12 \), \( b = -200 \), and \( c = 616 \).Calculating:\[ x = \frac{200 \pm \sqrt{(-200)^2 - 4 \(12\) \(616\)}}{2 \(12\)} \]\[ x = \frac{200 \pm \sqrt{40000 - 29568}}{24}\]\[ x = \frac{200 \pm \sqrt{10432}}{24} \]\[ x = \frac{200 \pm 102.13}{24} \]\[ x \approx 12.59 \, \text{cm}\] or \[ x \approx 3.41 \, \text{cm}\]

Step 6: Evaluate the Critical Points

Substitute potential values of \( x \) into the volume equation to determine which gives the largest volume. Since \( x = 12.59 \, \text{cm} \) results in negative length or width, only \( x = 3.41 \, \text{cm} \) should be considered. Calculate and verify the volume is maximized at this \( x \) value.

Step 7: Consider the Constraints

Ensure that the choice for \( x \) is within feasible constraints, i.e., \( 0 < x < 11 \) because length or width cannot be negative. Taking \( x = 3.41 \) cm is reasonable within these constraints.

Key Concepts

Cubic Functions

Cubic functions are polynomial functions of degree three. In this context, our volume equation is a cubic function:

• It has the general form: \( ax^3 + bx^2 + cx + d \).
• The equation for the box's volume simplifies to \( V = 4x^3 - 100x^2 + 616x \).
• This function helps us calculate the box's volume based on the size \( x \).

Cubic functions are very important in calculus for modeling real-world scenarios involving rates of change. They often have turning points where they change direction, which is crucial in optimization problems like this one.
To find the optimal solution, we need to identify these turning points, also known as critical points.

Derivative of a Function

Derivatives are fundamental when attempting to find critical points, helping us understand how the function behaves. The derivative of a function provides insight into how the function's output changes with respect to its input. In our problem, the derivative of the volume function, \( V' = 12x^2 - 200x + 616 \), is key to finding the box size that maximizes volume:

• Taking the derivative allows us to find slopes of tangent lines to the curve of the volume function.
• By setting the derivative equal to zero, \( V' = 0 \), we can find possible critical points, where the slope of the tangent is zero.
• These points indicate where the volume stops increasing or decreasing.

This process is called finding the critical points, and it's a cornerstone of optimization in calculus.

Quadratic Equation

To solve for the critical points, we need to use what's called a quadratic equation. Here, we've got a derivative quadratic equation \( 12x^2 - 200x + 616 = 0 \). This equation is solved using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 12 \), \( b = -200 \), and \( c = 616 \).
• Plug in the values to find \( x \) and you get approximate values of 12.59 cm and 3.41 cm.

The quadratic formula is a powerful tool in algebra, allowing us to find the roots of any quadratic equation, which represent the potential solutions for maximizing or minimizing a function, in this case, the volume of the box.

Critical Points

Critical points are where the function's derivative is zero, indicating potential maxima or minima. For the box problem, these points represent potential sizes for the square cuts that might maximize the box volume. After solving the quadratic equation, the values of \( x \) found are 12.59 cm and 3.41 cm:

• However, we must consider practical constraints. The length \( 22 - 2x \) and width \( 28 - 2x \) must remain positive.
• This restricts feasible \( x \) values to between 0 and 11 cm.
• Therefore, only \( x = 3.41 \) cm is a practical solution, while 12.59 cm would make dimensions negative.

By checking these constraints, we confirm which critical point is valid, ensuring our solution is both mathematically and practically sound.