Let $ u = e^{-0.5t} $, so $ f(t) = \frac{u}{(1 + 9u)^2} $. Now find the maximum of this function with respect to $ u > 0 $.

["Optimizing the Function $ f(u) = \frac{u}{(1 + 9u)^2} $: Finding Its Maximum for $ u > 0 $", "When solving optimization problems in applied mathematics and engineering, one common strategy is to transform variables to simplify complex expressions. Here, we are given $ f(t) = \frac{u}{(1 + 9u)^2} $ with $ u = e^{-0.5t} $, so $ u > 0 $. Since $ t $ varies over all real numbers, $ u $ spans $ (0, \infty) $. The goal is to find the maximum value of $ f(u) $ with respect to $ u > 0 $, streamlining analysis by focusing on a single variable.", "---", "### Step 1: Reformulate the Objective Function", "We simplify the optimization by replacing $ u $ with a single variable:", "$$\nf(u) = \frac{u}{(1 + 9u)^2}, \quad u > 0\n$$", "Our task is to compute $ \max_{u > 0} f(u) $.", "---", "### Step 2: Use Calculus to Find Critical Points", "To find maxima, compute the derivative $ f'(u) $ and solve $ f'(u) = 0 $.", "Let $ f(u) = \frac{u}{(1 + 9u)^2} $. Use the quotient rule:", "If $ f(u) = \frac{N(u)}{D(u)} $, then\n$$\nf'(u) = \frac{N' D - N D'}{D^2}\n$$", "Here,\n- $ N(u) = u \Rightarrow N'(u) = 1 $\n- $ D(u) = (1 + 9u)^2 \Rightarrow D'(u) = 2(1 + 9u)(9) = 18(1 + 9u) $", "Now compute:", "$$\nf'(u) = \frac{(1)(1 + 9u)^2 - u \cdot 18(1 + 9u)}{(1 + 9u)^4}\n$$", "Factor $ (1 + 9u) $ in the numerator:", "$$\nf'(u) = \frac{(1 + 9u)\left[(1 + 9u) - 18u\right]}{(1 + 9u)^4} = \frac{(1 + 9u - 18u)}{(1 + 9u)^3} = \frac{1 - 9u}{(1 + 9u)^3}\n$$", "---", "### Step 3: Solve $ f'(u) = 0 $", "Set numerator equal to zero:", "$$\n1 - 9u = 0 \Rightarrow u = \frac{1}{9}\n$$", "Since $ u > 0 $, this critical point is valid.", "---", "### Step 4: Confirm It Is a Maximum", "Analyze the sign of $ f'(u) $:", "- For $ 0 < u < \frac{1}{9} $, $ 1 - 9u > 0 \Rightarrow f'(u) > 0 $: increasing\n- For $ u > \frac{1}{9} $, $ 1 - 9u < 0 \Rightarrow f'(u) < 0 $: decreasing", "Thus, $ f(u) $ increases to $ u = \frac{1}{9} $, then decreases — confirming a maximum at $ u = \frac{1}{9} $.", "---", "### Step 5: Compute the Maximum Value", "Substitute $ u = \frac{1}{9} $ into $ f(u) $:", "$$\nf\left(\frac{1}{9}\right) = \frac{\frac{1}{9}}{\left(1 + 9 \cdot \frac{1}{9}\right)^2} = \frac{\frac{1}{9}}{(1 + 1)^2} = \frac{\frac{1}{9}}{4} = \frac{1}{36}\n$$", "---", "### Conclusion", "The maximum value of $ f(t) = \frac{u}{(1 + 9u)^2} $ for $ u > 0 $ occurs at $ u = \frac{1}{9} $, and the maximum value is $ \frac{1}{36} $. By transforming the original function using $ u = e^{-0.5t} $, we reduced the problem to a single-variable optimization — a powerful technique for real-world models in growth, decay, and constrained systems.", "This method illustrates how variable substitution can reveal key features of complex functions, making it easier to identify maxima, minima, and behavior over domains. Whether in physics, economics, or data science, such transformations remain essential tools in applied calculus.", "---", "Keywords: optimize $ f(u) = \frac{u}{(1 + 9u)^2} $, maximum of function $ u > 0 $, calculus optimization, variable substitution, exponential transformation, derivative test, $ u = e^{-0.5t} $ model."]