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The Ultimate Crash Course for STEM Majors · Sample Lesson
Calculus Optimization – Find Two Positive Numbers with Product 100 and Minimum Sum
By Author Jonathan David
How do you find two positive numbers whose product is 100 and whose sum is as small as possible? This differential calculus optimization problem illustrates how to identify a constraint, build a function of one variable, and use derivatives to locate a minimum.
Question: Find two positive numbers whose product is 100 and whose sum is a minimum.
Answer: The numbers are $10$ and $10$, giving a minimum sum of $20$.
Identify the Constraint and the Objective
Let the two positive numbers be $x$ and $y$. The phrase “whose product is 100” supplies the constraint:
The phrase “whose sum is a minimum” identifies the quantity to optimize:
Keep these roles separate. The product stays fixed while the sum changes as you choose different pairs of positive numbers.
Write the Sum as a Function of One Variable
Because $x$ is positive, it is nonzero. Solve the constraint for $y$:
Substitute this expression into the sum:
The domain $x>0$ comes from the requirement that both numbers be positive. It will determine which algebraic candidates are acceptable.
Differentiate and Find the Critical Point
Differentiate the sum with respect to $x$:
This is an optimization problem, so the derivative is taken with respect to $x$. Differentiating with respect to time is a related rates technique and is not needed here.
To find a stationary point, set the derivative equal to zero. Writing it as a single fraction makes the numerator and denominator restrictions clear:
A fraction equals zero when its numerator is zero and its denominator is nonzero. Since this problem requires $x>0$, reject $x=-10$. The only critical point in the domain is $x=10$.
Domain reminder: Multiplying the equation by $x^2$ is also valid once $x\ne0$ has been established. The essential habit is to preserve the domain restrictions throughout the algebra. Here, $x=0$ is excluded from the original function.
Verify That the Sum Is a Minimum
Finding a critical point gives a candidate. To establish the minimum, examine the sign of the derivative:
- For $0<x<10$, $S'(x)<0$, so the sum is decreasing.
- For $x>10$, $S'(x)>0$, so the sum is increasing.
The function decreases up to $x=10$ and increases afterward. Therefore, $x=10$ gives the absolute minimum over the positive domain. This verification expands on the calculations visible in the sample page.
Find the Second Number and Check the Answer
Return to the constraint to find $y$:
The two positive numbers are $10$ and $10$, and their minimum possible sum is $20$.
The Rectangle Interpretation
Think of $x$ and $y$ as the side lengths of a rectangle. The constraint $xy=100$ fixes its area at 100 square units. Its perimeter is $P=2(x+y)$, so minimizing the sum also minimizes the perimeter.
The minimizing rectangle has equal side lengths: it is a square measuring 10 units by 10 units, with a perimeter of 40 units.
The study method is to identify the constraint, build the objective function, state its domain, differentiate, verify the extremum, and check the final answer against the original question.
Original Page from Book
