Optimization Problems in Calculus

The primary goal of optimization is to determine the optimum values, which refers to either the absolute minimum or absolute maximum of a function within a real-life context.

Step 1: Identify Quantities. List all given quantities and those that need to be determined. Creating a sketch is recommended for visualization.
Step 2: Write a Primary Equation. Formulate an equation for the specific quantity that must be maximized or minimized.
Step 3: Reduce Variables. If the primary equation has more than one independent variable, use a secondary equation to relate the variables and reduce the primary equation to a single independent variable.
Step 4: Determine the Feasible Domain. Define the interval of values for the independent variable that makes sense within the physical constraints of the problem.
Step 5: Apply Calculus. Use differentiation and optimization tests (such as the First-Derivative Test) to find the desired maximum or minimum value.

Problem: Design an open box with a square base and a surface area $S$ of $108 \, \text{square inches}$ to achieve maximum volume.
Primary Equation (Volume): $V = x^2 h$
Secondary Equation (Surface Area): $S = x^2 + 4xh = 108$
Reduction: Solving the secondary equation for $h$ gives $h = \frac{108 - x^2}{4x}$ . Substituting this into the volume formula yields a function of one variable: $V = x \left( \frac{108 - x^2}{4} \right) = 27x - \frac{x^3}{4}$ .
Feasible Domain: Since $x$ must be positive and $x^2 \le 108$ , the domain is approximately 0 < x \le \sqrt{108}.
Result: The maximum volume occurs when $x = 6 \, \text{inches}$ and $h = 3 \, \text{inches}$ .

Problem: A rectangular page must contain $24 \, \text{square inches}$ of print with margins of $1.5 \, \text{inches}$ at the top/bottom and $1 \, \text{inch}$ on each side. Minimize the total paper area $A$ .
Primary Equation: $A = (x + 3)(y + 2)$
Secondary Equation (Print Area): $xy = 24 \Rightarrow y = \frac{24}{x}$
Function Reduction: $A = 2x + 30 + \frac{72}{x}$
Feasible Domain: x > 0
Critical Numbers: Differentiation leads to a critical number at $x = 6$ . Since $x = -6$ is not in the feasible domain, only the positive value is considered.
Result: Using the First-Derivative Test, the minimum area is confirmed at $x = 6$ . The final dimensions of the page include the margins.