Hemodialysis Hollow-Fiber Capsule Volume Maximization (Biomedical Engineering) Solution

Problem Overview

A biomedical engineering team designs an artificial kidney dialyzer housing modeled as a closed cylinder of fixed outer surface area $S_0 = 150\pi\text{ cm}^2$. We maximize internal dialyzer clearance volume $V = \pi r^2 h$, establish the closed physical domain, and evaluate extrema via the Closed Interval Method.

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Step-by-Step Mathematical Solution

Step 1: Formulate Governing Equations

The governing process follows the Applied Optimization Modeling Framework: (Defined in Applied Optimization Modeling Framework.)

1. Objective Function (Maximize Volume):
   $$V(r, h) = \pi r^2 h$$
   (Formula defined in Geometric Volume and Surface Area Formulas.)
2. Fixed Surface Area Constraint Equation: A closed right circular cylinder has two circular end manifold plates and one cylindrical barrel:
   $$S = 2\pi r^2 + 2\pi rh = 150\pi\text{ cm}^2$$
   Divide the entire constraint equation by $2\pi$:
   $$r^2 + rh = 75$$

Step 2: Variable Reduction (Substitution)

Isolate $h$ from the surface area constraint:

Substitute $h$ into the volume objective function:

Step 3: Establish the Context-Driven Domain

(Governed by Context-Driven Domain in Geometric Optimization.)

1. The radius cannot be negative: $r \ge 0$.
2. The housing length $h$ cannot be negative:
   $$h \ge 0 \implies \frac{75 - r^2}{r} \ge 0 \implies 75 - r^2 \ge 0 \implies r^2 \le 75$$
   $$r \le \sqrt{75} = 5\sqrt{3} \approx 8.66\text{ cm}$$
3. Feasible domain is the closed interval:
   $$r \in [0, \sqrt{75}]$$

Step 4: Differentiate and Find Critical Numbers

Differentiate $V(r)$ with respect to $r$ using the Power Rule:

Set $V'(r) = 0$ to identify stationary points: (Reject $r = -5\text{ cm}$ as non-physical). The critical number $r = 5\text{ cm}$ lies strictly within the interior of the domain $[0, \sqrt{75}]$ because $5 < \sqrt{75} \approx 8.66$.

Step 5: Execute Candidate Height Evaluation (Closed Interval Method)

Because $V(r)$ is a polynomial continuous on the closed bounded interval $[0, \sqrt{75}]$, the Extreme Value Theorem (defined in Extreme Value Theorem) guarantees the existence of an absolute maximum.

Evaluate $V(r)$ at both boundary endpoints and the interior critical point:

1. Left Boundary ($r = 0\text{ cm}$ - degenerate wire):
   $$V(0) = \pi (75(0) - (0)^3) = 0\text{ cm}^3$$
2. Right Boundary ($r = \sqrt{75}\text{ cm}$ - zero-length flat disk):
   $$V(\sqrt{75}) = \pi (75\sqrt{75} - (\sqrt{75})^3) = \pi (75\sqrt{75} - 75\sqrt{75}) = 0\text{ cm}^3$$
3. Interior Critical Point ($r = 5\text{ cm}$):
   $$V(5) = \pi (75(5) - (5)^3) = \pi (375 - 125) = 250\pi\text{ cm}^3 \approx 785.4\text{ cm}^3$$

Comparing candidate values $\{0, 0, 250\pi\}$, the absolute maximum internal volume is $250\pi\text{ cm}^3$ (or $785.4\text{ mL}$), occurring at $r = 5\text{ cm}$.

Step 6: Verify Optimal Aspect Ratio

Calculate the capsule length $h$:

Check the aspect ratio: The optimal cylinder length ($10\text{ cm}$) exactly equals its diameter ($2r = 10\text{ cm}$), confirming that whether maximizing volume for a fixed surface area or minimizing surface area for a fixed volume, the closed right cylinder universally optimizes at $h = 2r$ (as established in Optimal Aspect Ratios in Geometric Containers).