Question: A pharmacologist is modeling the concentration of a drug in the bloodstream over time with the function $ C(t) = \frac{3t}{t^2 + 4} $. Find the time $ t \geq 0 $ at which the concentration is maximized.

["Title: Maximizing Drug Concentration Defined by $ C(t) = \frac{3t}{t^2 + 4} $: A Pharmacological Analysis", "Meta Description:\nA pharmacologist models drug concentration in the bloodstream with $ C(t) = \frac{3t}{t^2 + 4} $. This article reveals how to find the time $ t \geq 0 $ at which concentration peaks using calculus.", "---", "# Maximizing Drug Concentration: Finding the Peak of $ C(t) = \frac{3t}{t^2 + 4} $", "Understanding drug efficacy relies heavily on understanding how a drug’s concentration changes over time in the bloodstream. Pharmacologists often model this relationship using mathematical functions, and one such key example is:", "$$\nC(t) = \frac{3t}{t^2 + 4}\n$$", "where $ C(t) $ represents drug concentration at time $ t $ (in hours), and $ t \geq 0 $. But when is this concentration maximized? This article shows how to determine the time $ t $ at which the drug concentration peaks by applying calculus.", "---", "## Why Maximizing Concentration Matters", "Drug concentration determines both therapeutic effect and risk of toxicity. The goal is to achieve optimal levels—high enough to be effective but low enough to minimize side effects. Identifying the time of maximum concentration helps clinicians predict peak plasma levels and schedule dosing accordingly.", "---", "## Step 1: Analyze the Function $ C(t) $", "We are given:\n$$\nC(t) = \frac{3t}{t^2 + 4}\n$$", "This is a rational function with a linear numerator and quadratic denominator. Since $ t \geq 0 $, we seek the global maximum on this domain.", "---", "## Step 2: Use Calculus to Find Critical Points", "To find the maximum, compute the first derivative $ C'(t) $ and solve $ C'(t) = 0 $.", "Apply the quotient rule: if $ C(t) = \frac{u(t)}{v(t)} $, then\n$$\nC'(t) = \frac{u'v - uv'}{v^2}\n$$", "Let:\n- $ u(t) = 3t $ → $ u'(t) = 3 $\n- $ v(t) = t^2 + 4 $ → $ v'(t) = 2t $", "Now compute:\n$$\nC'(t) = \frac{(3)(t^2 + 4) - (3t)(2t)}{(t^2 + 4)^2} = \frac{3t^2 + 12 - 6t^2}{(t^2 + 4)^2} = \frac{-3t^2 + 12}{(t^2 + 4)^2}\n$$", "---", "## Step 3: Solve $ C'(t) = 0 $", "Set the numerator equal to zero:\n$$\n-3t^2 + 12 = 0 \quad \Rightarrow \quad 3t^2 = 12 \quad \Rightarrow \quad t^2 = 4 \quad \Rightarrow \quad t = 2 \quad (\ ext{since } t \geq 0)\n$$", "We discard $ t = -2 $ because time cannot be negative.", "---", "## Step 4: Confirm It’s a Maximum", "Check the sign of $ C'(t) $ around $ t = 2 $:", "- For $ t < 2 $, say $ t = 1 $: $ -3(1)^2 + 12 = 9 > 0 $ → $ C'(t) > 0 $: increasing\n- For $ t > 2 $, say $ t = 3 $: $ -3(9) + 12 = -15 < 0 $ → $ C'(t) < 0 $: decreasing", "Since the function changes from increasing to decreasing at $ t = 2 $, this is a local (and global) maximum on $ [0, \infty) $.", "---", "## Step 5: Interpret the Result", "At $ t = 2 $ hours, the drug concentration reaches its peak value. Substituting back:", "$$\nC(2) = \frac{3(2)}{(2)^2 + 4} = \frac{6}{4 + 4} = \frac{6}{8} = 0.75 \ ext{ mg/L (or concentration units)}\n$$", "---", "## Conclusion", "By modeling drug concentration using $ C(t) = \frac{3t}{t^2 + 4} $, pharmacologists can mathematically identify the optimal time to reach maximum therapeutic levels. Using calculus, we found that the concentration peaks at:", "$$\n\boxed{2 \ ext{ hours}}\n$$", "This insight supports better dosing regimens, improved patient outcomes, and personalized treatment plans based on pharmacokinetic modeling.", "---", "Keywords: drug concentration model, pharmacokinetics, calculus application, peak drug level, time to maximum concentration, $ C(t) = \frac{3t}{t^2 + 4} $, maximize $ C(t) $, Raleigh-Phuoc model insight", "For more on modeling drug kinetics: subscribe to our pharmacology newsletter or consult a clinical pharmacologist."]