Solution: To maximize $ C(t) = \frac{3t}{t^2 + 4} $, we take its derivative using the quotient rule:
["Maximizing $ C(t) = \dfrac{3t}{t^2 + 4} $: A Step-by-Step Solution Using the Quotient Rule", "In optimization problems, one of the most powerful tools is calculus—specifically, finding critical points by taking the derivative. When maximizing rational functions like $ C(t) = \dfrac{3t}{t^2 + 4} $, the quotient rule provides a precise way to compute the first derivative and identify where the function reaches its maximum.", "This article explains how to maximize $ C(t) $ by applying the quotient rule, highlighting key calculus concepts and real-world applications.", "---", "### What is $ C(t) $?", "The function\n$$\nC(t) = \frac{3t}{t^2 + 4}\n$$\nmodels various phenomena where a linear input $ 3t $ interacts with a quadratic resistance or growth term $ t^2 + 4 $. Such functions often appear in engineering, economics, and physics—especially in optimization scenarios.", "Our goal:\nFind the value of $ t $ that maximizes $ C(t) $, using the quotient rule to compute $ C'(t) $.", "---", "### Step 1: Recall the Quotient Rule", "For a function $ C(t) = \frac{u(t)}{v(t)} $, the derivative is:\n$$\nC'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}\n$$", "Here,\n- $ u(t) = 3t $ → $ u'(t) = 3 $\n- $ v(t) = t^2 + 4 $ → $ v'(t) = 2t $", "---", "### Step 2: Apply the Quotient Rule", "Substitute into the formula:\n$$\nC'(t) = \frac{(3)(t^2 + 4) - (3t)(2t)}{(t^2 + 4)^2}\n$$", "Now simplify the numerator:\n$$\n3(t^2 + 4) - 6t^2 = 3t^2 + 12 - 6t^2 = -3t^2 + 12\n$$", "So,\n$$\nC'(t) = \frac{-3t^2 + 12}{(t^2 + 4)^2}\n$$", "---", "### Step 3: Find Critical Points", "Maximum or minimum values occur where $ C'(t) = 0 $ or is undefined. Since the denominator $ (t^2 + 4)^2 > 0 $ for all real $ t $, $ C'(t) $ is defined everywhere.", "Set the numerator equal to zero:\n$$\n-3t^2 + 12 = 0\n\quad \Rightarrow \quad\nt^2 = 4\n\quad \Rightarrow \quad\nt = \pm 2\n$$", "---", "### Step 4: Determine the Maximum", "To confirm which critical point yields a maximum, examine the sign or evaluate $ C(t) $ at $ t = 2 $ and $ t = -2 $:", "- $ C(2) = \dfrac{3 \cdot 2}{2^2 + 4} = \dfrac{6}{8} = \dfrac{3}{4} = 0.75 $\n- $ C(-2) = \dfrac{3 \cdot (-2)}{(-2)^2 + 4} = \dfrac{-6}{8} = -0.75 $", "Clearly, $ C(t) $ is maximized at $ t = 2 $.", "---", "### Why This Matters", "This method—using the quotient rule to find $ C'(t) $—proves essential for optimizing rational functions. Real-world applications include:", "- Profit maximization, where revenue ($ 3t $) is divided by cost structures ($ t^2 + 4 $).\n- Efficiency modeling, analyzing output versus resource allocation.\n- Physics, such as optimizing signal responses in circuit analysis.", "---", "### Conclusion", "Maximizing $ C(t) = \dfrac{3t}{t^2 + 4} $ using the quotient rule is straightforward: differentiate with $ u = 3t $, $ v = t^2 + 4 $, apply $ C'(t) = \dfrac{u'v - uv'}{v^2} $, solve $ C'(t) = 0 $, and verify maximum values. The optimal solution occurs at $ t = 2 $, where $ C(t) $ reaches its peak at $ 0.75 $.", "Mastering the quotient rule empowers you to solve complex optimization problems efficiently—opening doors across science, engineering, and economics.", "---", "Keywords: $ C(t) = \dfrac{3t}{t^2 + 4} $, maximize function, quotient rule, calculus tutorial, optimization, derivative, rational function, real-world application, math problem solving.", "Meta Description: Learn how to maximize $ C(t) = \dfrac{3t}{t^2 + 4} $ by applying the quotient rule. Step-by-step guide with critical points and real-world relevance."]
