Problem 11 - Essential Calculus Early Transcendentals

Question

If 1200 \(\mathrm{cm}^{2}\) of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

Step-by-Step Solution

Answer

The largest possible volume of the box is \( 4000 \mathrm{cm}^3 \).

Step 1: Define the variables

Let the side of the square base be \( x \) and the height of the box be \( h \). The area of the base is \( x^2 \) and the area of the four sides is \( 4xh \). The box has no top, so the total surface area is \( x^2 + 4xh = 1200 \, \mathrm{cm}^2 \).

Step 2: Express height in terms of side length

Rearrange the surface area equation to solve for \( h \) in terms of \( x \):\[ h = \frac{1200 - x^2}{4x} \]

Step 3: Write the volume function

The volume \( V \) of the box is given by the formula \( V = x^2h \). Substitute \( h \) from Step 2 into the volume formula:\[ V = x^2 \cdot \frac{1200 - x^2}{4x} = \frac{x(1200 - x^2)}{4} \]

Step 4: Simplify the volume function

Simplify the expression for the volume:\[ V = \frac{1200x - x^3}{4} \]

Step 5: Find the derivative of the volume function

To find the maximum volume, find the derivative of \( V \), denoted \( V' \), and set it equal to zero. First, differentiate:\[ V' = \frac{d}{dx} \left( \frac{1200x - x^3}{4} \right) = \frac{1200 - 3x^2}{4} \]

Step 6: Solve for critical points

Set \( V' = 0 \) to find the critical points:\[ \frac{1200 - 3x^2}{4} = 0 \]\[ 1200 - 3x^2 = 0 \]\[ 3x^2 = 1200 \]\[ x^2 = 400 \]\[ x = 20 \]

Step 7: Determine the height corresponding to the critical point

Substitute \( x = 20 \) back into the equation for \( h \) from Step 2:\[ h = \frac{1200 - 20^2}{4 \times 20} = \frac{1200 - 400}{80} = \frac{800}{80} = 10 \]

Step 8: Calculate the volume at the critical point

Now, calculate the volume when \( x = 20 \) and \( h = 10 \):\[ V = x^2h = 20^2 \times 10 = 400 \times 10 = 4000 \, \mathrm{cm}^3 \]

Step 9: Verify maximum volume using second derivative test

The second derivative of \( V \), denoted \( V'' \), is:\[ V'' = \frac{d^2}{dx^2} \left( \frac{1200x - x^3}{4} \right) = \frac{-6x}{4} = -\frac{3x}{2} \]Substitute \( x = 20 \) into \( V'' \) to confirm:\[ V''(20) = -\frac{3(20)}{2} = -30 \]Since \( V'' < 0 \), the value at \( x = 20 \) is a maximum.

Key Concepts

Critical Points

In calculus optimization problems, critical points are crucial for finding maxima or minima of a function. A critical point occurs where the derivative of the function equals zero or where the derivative does not exist. These points are potential candidates for local maxima or minima.
To solve the problem of maximizing the box's volume, we first defined the volume function in terms of the base side length, \( x \). Our task in the step-by-step solution was to identify where this function reaches its maximum value.
By differentiating the volume function \( V(x) \) and setting \( V'(x) = 0 \), we solved for \( x \) to find critical points. Critical points are valid only if they fit within the constraints of the problem, such as those set by the surface area condition, guiding us to find \( x = 20 \) as a valid critical point.

Derivative

Derivatives are essential for understanding the behavior of functions and play a key role in optimization. They measure the rate at which a function's value changes with respect to changes in its input value.
In our example, we derived the volume function \( V(x) = \frac{1200x - x^3}{4} \) with respect to \( x \) to get the first derivative \( V'(x) \). The derivative \( V'(x) \) provides insights into where the function is increasing or decreasing, helping to locate critical points where it could potentially achieve its maximum.

• Finding the critical points involves solving \( V'(x) = 0 \).
• In this problem, this set us up to find when the change in the volume is zero, which corresponds to a maximum volume point.

Derivatives like \( V'(x) \) are integral to these optimization tasks.

Surface Area Constraint

Constraints in optimization problems restrict the values variables can take. For this exercise, we had a fixed surface area of \( 1200 \ \mathrm{cm}^2 \) for the box, forming an essential constraint.
The surface area, expressed as \( x^2 + 4xh = 1200 \), includes the square base and four sides but excludes a top. This constraint helped us express the height \( h \) in terms of the base side \( x \):
\[ h = \frac{1200 - x^2}{4x} \]
By substituting this derived relationship into the volume equation, we ensured that our volume function accounted for this constraint. Understanding constraints is fundamental because it:

• Guides both the derivation and evaluation of functions.
• Ensures that solutions deal with real-world limitations, such as material availability as seen here.

The constraint solidifies the feasibility of the calculated solutions.

Volume Maximization

The main aim of optimization in this context was to find the maximum volume of the box given the constraints. Volume maximization combines finding critical points and using calculus to ensure that these are maxima, rather than minima or saddle points.
With calculated critical points, we exercised the second derivative test, using \( V''(x) \) to confirm that the critical point \( x = 20 \) is indeed a maximum. Here, \( V''(20) = -30 \), showing that the slope is decreasing, confirming a local maximum.

• The second derivative test checks concavity to ensure a point is a maximum when \( V''(x) < 0 \).
• With this, the box's calculated maximum volume for \( x = 20 \) and \( h = 10 \) becomes \( 4000 \ \mathrm{cm}^3 \).

Using calculus tools like the derivative and second derivative tests effectively achieves reliable solutions for maximizing variables under specific conditions.