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
Second Derivative Test – Verifying a Minimum and Introducing Box Optimization
By Author Jonathan David
Finding a critical point is one stage of a calculus optimization problem. The next stage is to determine whether that point gives a minimum, a maximum, or neither. This sample continues the problem of finding two positive numbers whose product is 100 and whose sum is as small as possible.
The page uses the second derivative test to classify the critical point, interprets the result graphically, and introduces a new problem involving the maximum volume of an open-top box.
Review the Function Being Minimized
Let the positive numbers be $x$ and $y$. Their product is fixed:
Differentiating and setting the derivative equal to zero gives:
Negative values and zero are excluded because this particular question requires positive numbers. Optimization problems can have other domains, so always obtain restrictions from the question and the mathematical model.
Apply the Second Derivative Test
At a critical point $c$ where $f'(c)=0$, a positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum. If $f”(c)=0$, the test is inconclusive.
For the sum function, differentiate again:
Since $S'(10)=0$ and $S”(10)>0$, the second derivative test establishes a local minimum at $x=10$.
Explain Why This Is the Absolute Minimum
The test at a single point classifies a local extremum. To show that this result is the smallest sum over the entire positive domain, examine the first derivative:
- For $0<x<10$, $S'(x)<0$, so $S$ decreases.
- For $x>10$, $S'(x)>0$, so $S$ increases.
The sum decreases toward $x=10$ and increases afterward. Therefore, this point gives the absolute minimum for $x>0$.
The two numbers are $10$ and $10$, and the minimum sum is $20$. Checking the product gives $10\cdot10=100$, as required.
Read the Minimum on the Graph
The original page shows the graph of the sum function. Its lowest point on the positive domain is:
The horizontal coordinate, $10$, represents the first number. The vertical coordinate, $20$, represents the sum. The second number is also $10$, but it is not the vertical coordinate on this graph.
Keeping the input and output distinct helps you explain the conclusion correctly: the minimum occurs at $x=10$, and the minimum function value is $20$.
Next Problem – Maximize the Volume of an Open-Top Box
Question: If $1200\text{ cm}^2$ of material is available to make a box with a square base and an open top, find the largest possible volume.
Answer shown on the original page: $4000\text{ cm}^3$.
The page introduces this next exercise without showing its full derivation. The following setup expands on that introduction.
Let $x$ be the side length of the square base and $h$ be the height, both measured in centimeters. The box has one base and four rectangular sides. Because it has no top, the material constraint is:
Its volume is $V=x^2h$. Solve the material constraint for the height and substitute into the volume:
Positive dimensions require $0<x<\sqrt{1200}$. A complete optimization solution would differentiate $V(x)$, identify the relevant critical point, and verify the maximum. The dimensions $x=20$ and $h=10$ match the stated result:
A Repeatable Method for Calculus Optimization
Both exercises use the same core process: identify the quantity being optimized, write the constraint, reduce the objective to one variable, determine the domain, and use derivatives to justify the result.
Finish by answering the original question in context. Distinguish the dimensions or numbers you found from the minimum or maximum value they produce, and include units where appropriate.
Original Page from Book
