Sample preview: The original book page is an image. This written expansion explains part of that page to provide a readable sample of the lesson. It is not the complete lesson or book.
The Ultimate Crash Course for STEM Majors · Sample Lesson
Open-Top Box Optimization – Build the Volume Function and Find the Critical Point
By Author Jonathan David
An open-top box optimization problem connects geometry with differential calculus. The geometry supplies the surface area constraint and volume formula. Calculus then helps determine which dimensions produce the largest volume.
This sample develops the model for a box with a square base made from $1200\text{ cm}^2$ of material. It shows how to reduce the volume to a function of one variable and find a critical point.
Define the Dimensions Before Writing Equations
Let $x$ represent the side length of the square base and $y$ represent the height, both measured in centimeters. Because the base is square, the box has length $x$, width $x$, and height $y$.
- Base area: $x^2$.
- Area of each side: $xy$.
- Combined area of four sides: $4xy$.
- Volume: $x^2y$.
The units help distinguish the quantities: square centimeters measure area, while cubic centimeters measure volume.
Write the Surface Area Constraint
The box has one square base and four rectangular sides. Its top is open, so no top panel is included in the material calculation.
Read the question carefully: The original illustration shows a raised lid, but the stated problem specifies an open top. Follow the problem’s conditions and omit the lid’s area.
A closed box would include another $x^2$ for the top. That would change the constraint and produce a different optimization problem.
Identify the Quantity to Maximize
The available material fixes the surface area. The quantity we want to maximize is the volume:
At this stage, the volume depends on two variables. Use the surface area constraint to express the height in terms of the base length.
Build a Volume Function of One Variable
Solve the constraint for $y$, remembering that a physical base length must satisfy $x>0$:
Substitute this expression into $V=x^2y$ and simplify:
The volume now depends only on $x$. Each allowed base length determines a corresponding height through the material constraint.
Determine the Physical Domain
Both the base length and the height must be positive. Since $x>0$, the height formula requires $1200-x^2>0$. Therefore:
Although the simplified polynomial can be evaluated outside this interval, those values do not describe a box with positive dimensions under the stated constraint.
Differentiate and Find the Critical Point
Differentiate the volume function with respect to the base length:
Set the derivative equal to zero to locate a candidate for the maximum:
Reject $x=-20$ because a side length cannot be negative. The remaining candidate, $x=20$, lies within the physical domain.
What Remains to Complete the Optimization?
The original page ends by identifying $x=20$ as a possible extremum and introducing the second derivative test. A critical point alone does not establish a maximum; it must be classified and checked over the relevant domain.
As an expansion of that next step, the second derivative is:
This establishes a local maximum. Moreover, $V'(x)$ is positive before $20$ and negative after $20$ within the physical domain, so the volume increases and then decreases. Thus, this critical point gives the absolute maximum.
Substituting the base length into the height and volume formulas completes the result:
The maximizing box has a 20 cm by 20 cm base and a height of 10 cm. The central lesson is to build the correct geometric model before differentiating, then justify why the candidate gives the requested maximum.
Original Page from Book
