Steps for Solving Maxima/Minima Word Problems
📌 The first step involves drawing a diagram if necessary to analyze the problem setup.
📝 Next, write an equation representing the quantity to be maximized or minimized, typically involving two or more variables.
🧮 Use relationships between variables to express the equation as a function of a single variable.
➗ Differentiate the single-variable function and equate the derivative to zero to find critical points.

Example 1: Maximizing $x^2y$ with $x+y=300$
📐 The constraint equation is $x+y=300$, leading to $y=300-x$.
📈 The function to maximize is $f(x) = x^2(300-x) = 300x^2 - x^3$.
🧮 The first derivative is $f'(x) = 600x - 3x^2$. Setting $f'(x)=0$ yields roots $x=0$ and $x=200$.
✔️ The second derivative test ($f''(x) = 600 - 6x$) confirms a relative maximum at $x=200$ ($f''(200) = -600$).
🔢 The two non-negative numbers are $x=200$ and $y=300-200=100$.

Example 2: Maximizing Volume of an Open Box
📦 The dimensions of the base are $(8-2x)$ and $(3-2x)$, and the height is $x$.
🔺 The volume function is $V(x) = x(8-2x)(3-2x)$, which simplifies to $V(x) = 4x^3 - 22x^2 + 24x$.
➗ The first derivative is $V'(x) = 12x^2 - 44x + 24$. Setting $V'(x)=0$ and simplifying gives $3x^2 - 11x + 6 = 0$.
✂️ The critical values are $x=\frac{2}{3}$ and $x=3$. Evaluating the second derivative $V''(x) = 24x - 44$ shows that $x=\frac{2}{3}$ yields a maximum volume ($V''(\frac{2}{3}) = -28$).

Example 3: Maximizing Area of a Rectangle with Fixed Perimeter
🔗 Given a perimeter of 16 feet, $2x + 2y = 16$, so $x+y=8$, or $y=8-x$.
🔺 The area function to maximize is $A(x) = x(8-x) = 8x - x^2$.
📏 Setting the derivative $A'(x) = 8 - 2x$ to zero gives the critical point $x=4$.
🔄 The second derivative $A''(x) = -2$ confirms a maximum area at $x=4$.
📐 The dimensions that maximize the area are $x=4$ feet and $y=4$ feet (a square).

Key Points & Insights
➡️ Always use the second derivative test to definitively determine if a critical point corresponds to a relative maximum ($f''(c) < 0$) or minimum ($f''(c) > 0$).
➡️ In multi-variable optimization problems, the initial crucial step is reducing the objective function to depend on a single variable using the constraint equation.
➡️ For problems involving physical dimensions, ensure that the resulting critical points are physically plausible (e.g., dimensions cannot be negative or zero unless trivial).

📸 Video summarized with SummaryTube.com on Nov 25, 2025, 13:48 UTC