To find when \(C(t) = \frac{kt}{t^2 + 1}\) is maximized, find the critical points by setting the derivative \(C'(t)\) to zero.
["Finding the Time When ( C(t) = \frac{kt}{t^2 + 1} ) Peaks: A Step-by-Step Guide to Maximizing the Function", "Mathematical modeling often involves identifying when a function reaches its maximum — especially in real-world applications like physics, economics, and biology. One such function commonly studied is:", "[\nC(t) = \frac{kt}{t^2 + 1}\n]", "where ( k ) is a positive constant, and ( t \geq 0 ). In this article, we’ll explore how to find when ( C(t) ) is maximized by finding its critical points — specifically, by solving ( C'(t) = 0 ).", "---", "### Why Maximize ( C(t) )?", "This function typically arises when modeling quantities like concentration, efficiency, or response that increase initially but level off due to a quadratic denominator. Understanding its maximum provides insights into optimal timing in dynamic systems.", "---", "### Step 1: Compute the Derivative ( C'(t) )", "To find the maximum, first compute the derivative of ( C(t) ) with respect to ( t ). Since ( C(t) ) is a quotient of two functions, we apply the quotient rule:", "[\nC'(t) = \frac{(k)(t^2 + 1) - kt(2t)}{(t^2 + 1)^2}\n]", "Simplify the numerator:", "[\nk(t^2 + 1) - 2kt^2 = kt^2 + k - 2kt^2 = -kt^2 + k\n]", "So,", "[\nC'(t) = \frac{k(1 - t^2)}{(t^2 + 1)^2}\n]", "---", "### Step 2: Find Critical Points by Setting ( C'(t) = 0 )", "Critical points occur where the derivative is zero (provided the denominator is not zero, which it never is since ( t^2 + 1 \geq 1 )).", "Set the numerator equal to zero:", "[\nk(1 - t^2) = 0\n]", "Since ( k <br/>\ne 0 ), we solve:", "[\n1 - t^2 = 0 \quad \Rightarrow \quad t^2 = 1 \quad \Rightarrow \quad t = \pm 1\n]", "But since ( t \geq 0 ) in context, we take only the positive solution:", "[\nt = 1\n]", "---", "### Step 3: Confirm It’s a Maximum", "To verify ( t = 1 ) is a maximum, examine the sign of ( C'(t) ) around ( t = 1 ):", "- For ( 0 \leq t < 1 ): ( 1 - t^2 > 0 ) ⇒ ( C'(t) > 0 ) → ( C(t) ) increasing\n- For ( t > 1 ): ( 1 - t^2 < 0 ) ⇒ ( C'(t) < 0 ) → ( C(t) ) decreasing", "Thus, ( C(t) ) increases up to ( t = 1 ) and decreases afterward, confirming a maximum at ( t = 1 ).", "---", "### Summary", "- The function ( C(t) = \frac{kt}{t^2 + 1} ) achieves its maximum at ( t = 1 )\n- Derivative: ( C'(t) = \frac{k(1 - t^2)}{(t^2 + 1)^2} )\n- Critical point: ( t = 1 ) (valid since ( t \geq 0 ))\n- Confirmed by sign analysis of ( C'(t) ), ( t = 1 ) is a maximum", "---", "### Final Notes", "Understanding when functions like ( C(t) ) peak is crucial for modeling and optimization. By computing derivatives, identifying critical points, and analyzing sign changes, you gain a systematic approach to finding maxima — a vital skill in calculus, engineering, and data science.", "For further reading, explore second derivative tests or optimization problems with other types of rational functions.", "---", "Keywords: maximize ( C(t) = \frac{kt}{t^2 + 1} ), find critical points, derivative ( C'(t) ), solve ( C'(t) = 0 ), maximize rational function, calculus optimization."]
