Question: A herpetologist models the population of a rare frog species in Madagascar with the function $ F(t) = \frac{100t}{t^2 + 4} $, where $ t $ is time in years. What is the maximum population growth rate?

["Understanding Maximum Population Growth Rate of an Endangered Frog Species Using Calculus", "Madagascar is home to a remarkable diversity of unique wildlife, including several endangered amphibian species. A recent scientific study models the population of a rare frog species using the function:", "$$\nF(t) = \frac{100t}{t^2 + 4}\n$$", "where $ t $ represents time in years since observation began. Understanding how this population evolves over time is critical for conservation efforts. An important question arises: What is the maximum population growth rate of this frog species?", "---", "### Modeling Growth: The Concept of Growth Rate", "The growth rate of a population function $ F(t) $ is given by its first derivative, $ F'(t) $, which represents how fast the population changes at any time $ t $. To find the maximum growth rate, we analyze $ F'(t) $ and determine its peak value.", "---", "### Step 1: Differentiate $ F(t) $", "We apply the quotient rule to differentiate:", "$$\nF(t) = \frac{100t}{t^2 + 4}\n$$", "Let:\n- $ u = 100t $ → $ u' = 100 $\n- $ v = t^2 + 4 $ → $ v' = 2t $", "Then:", "$$\nF'(t) = \frac{u'v - uv'}{v^2} = \frac{100(t^2 + 4) - (100t)(2t)}{(t^2 + 4)^2}\n$$", "Simplify the numerator:", "$$\n100(t^2 + 4) - 200t^2 = 100t^2 + 400 - 200t^2 = -100t^2 + 400\n$$", "So:", "$$\nF'(t) = \frac{-100t^2 + 400}{(t^2 + 4)^2}\n$$", "---", "### Step 2: Find Critical Points of $ F'(t) $", "To find the maximum of $ F'(t) $, we find where its derivative $ F''(t) = 0 $ (i.e., critical points of the growth rate).", "Rather than differentiate $ F'(t) $ directly (which becomes complex), we analyze the expression:", "$$\nF'(t) = \frac{-100t^2 + 400}{(t^2 + 4)^2}\n$$", "This is a rational function. The maximum occurs either at critical points or endpoints. Since $ t \geq 0 $, we focus on $ t \geq 0 $.", "Note:\n- At $ t = 0 $: $ F'(0) = \frac{400}{16} = 25 $\n- As $ t \ o \infty $, $ F'(t) \ o 0^- $ (approaches zero from negative side)", "To find the maximum, consider $ F'(t) $ as a function and use calculus: set $ F''(t) = 0 $. However, an efficient alternative is to exploit symmetry or substitute $ t = x $ and analyze $ F'(x) = \frac{400 - 100x^2}{(x^2 + 4)^2} $", "Let $ x^2 = u $, with $ u \geq 0 $. Then:", "$$\nF'(t) = \frac{400 - 100u}{(u + 4)^2}, \quad \ ext{where } u = t^2\n$$", "Define $ G(u) = \frac{400 - 100u}{(u + 4)^2} $", "Differentiate $ G(u) $ with respect to $ u $:", "$$\nG'(u) = \frac{(-100)(u+4)^2 - (400 - 100u) \cdot 2(u + 4)}{(u + 4)^4}\n$$", "Factor numerator:", "$$\n= \frac{-100(u+4) - 2(400 - 100u)}{(u + 4)^3} = \frac{-100u - 400 - 800 + 200u}{(u + 4)^3} = \frac{100u - 1200}{(u + 4)^3}\n$$", "Set $ G'(u) = 0 $:", "$$\n100u - 1200 = 0 \Rightarrow u = 12\n$$", "So maximum occurs at $ u = 12 $ → $ t = \sqrt{12} = 2\sqrt{3} $", "---", "### Step 3: Compute Maximum Growth Rate", "Substitute $ t = 2\sqrt{3} $ into $ F'(t) $:", "First compute numerator:", "$$\n-100t^2 + 400 = -100(12) + 400 = -1200 + 400 = -800\n$$", "Denominator:", "$$\n(t^2 + 4)^2 = (12 + 4)^2 = 16^2 = 256\n$$", "So:", "$$\nF'(2\sqrt{3}) = \frac{-800}{256} = -\frac{25}{8} = -3.125\n$$", "Wait — this is negative, but growth rate can be negative. However, we seek the maximum magnitude of growth — i.e., the largest rate of increase, not necessarily positive.", "But note: “maximum growth rate” often refers to the largest value of $ F'(t) $, even if negative in declining populations. However, in ecological contexts, we often interpret instantaneous rate of population increase — which here includes both rising and falling phases.", "But since the population starts increasing then decreasing (peak at $ t = 2\sqrt{3} $), the maximum rate of growth (i.e., steepest positive slope) occurs at this point — though absolute value of growth slows during decline.", "Thus, the maximum value of $ F'(t) $ is at $ t = 2\sqrt{3} $, and that value is:", "$$\nF'(2\sqrt{3}) = \frac{-800}{256} = -\frac{25}{8}\n$$", "But this is a minimum rate, not maximum growth.", "Wait — we confused: maximum growth rate means the largest positive value of $ F'(t) $.", "But from analysis, $ F'(t) $ starts at 25, decreases, reaches zero (extinction), and becomes negative. So the maximum growth rate (largest positive) occurs early, but we already found $ F'(t) $ decreases from 25 to 0.", "But earlier we found a critical point at $ t = 2\sqrt{3} $, and $ G'(u) = 0 $ at $ u=12 $, where $ F'(t) $ has a minimum, not maximum.", "So where is the maximum?", "Since $ F'(t) $ starts at 25 and decreases to 0, the maximum growth rate occurs at $ t = 0 $, where:", "$$\nF'(0) = \frac{-100(0)^2 + 400}{(0 + 4)^2} = \frac{400}{16} = 25\n$$", "But this is the initial rate — is this the maximum?", "Wait — could there be a maximum in $ F'(t) $ before $ t=0 $? No, $ t \geq 0 $.", "But is $ F'(t) $ increasing at $ t=0 $? Let’s analyze sign of $ F'(t) $:", "- For $ t > 0 $, numerator $ -100t^2 + 400 $ decreases from 400.\n- Denominator always positive.", "So $ F'(t) $ is decreasing for $ t > 0 $, starting at 25.", "Therefore, the maximum growth rate occurs at $ t = 0 $, with value 25 frogs per year per year — i.e., the population grows fastest in the first year.", "But is this realistic? Conservation biologists care about peak growth — and here, the population peaks early.", "But let’s verify using derivative test.", "We found $ G'(u) = \frac{100u - 1200}{(u+4)^3} $, which changes from positive to negative at $ u=12 $, so $ G(u) $ has a maximum at $ u=12 $, meaning $ F'(t) $ has a minimum, not a maximum.", "Since $ F'(t) $ starts at 25 and decreases monotonically, the maximum of $ F'(t) $ on $ [0, \infty) $ is at $ t = 0 $, and equals:", "$$\nF'(0) = \frac{400}{16} = 25\n$$", "But — is a growth rate of 25 frogs per year per year plausible?", "But let’s check the second derivative or plot behavior.", "Alternatively, we may have misunderstood: the maximum of the derivative $ F'(t) $ — but since $ F'(t) $ decreases from 25 to 0, the maximum is indeed at $ t = 0 $.", "However, in some contexts, “maximum growth rate” refers to the largest instantaneous increase, which is at the start.", "But perhaps the model predicts peak growth earlier — but our calculation shows $ F'(t) $ decreases.", "Wait — let’s evaluate $ F'(t) $ at small $ t $:", "- $ t = 0 $: $ F'(0) = 25 $\n- $ t = 1 $: $ F'(1) = \frac{-100 + 400}{(1 + 4)^2} = \frac{300}{25} = 12 $\n- $ t = 2 $: $ F'(2) = \frac{-400 + 400}{(4 + 4)^2} = 0 $\n- $ t = 3 $: $ F'(3) = \frac{-900 + 400}{(9 + 4)^2} = \frac{-500}{169} \approx -2.96 $", "So yes, $ F'(t) $ decreases from 25 to 0 — no maximum in between.", "Thus, the maximum value of the growth rate $ F'(t) $ is 25, occurring at $ t = 0 $.", "But this implies the population grows fastest in year zero — biologically plausible if reproduction is concentrated at birth.", "However, is this the maximum? Yes — it's the largest value.", "But let’s reconsider: could the maximum rate of change occur where $ F'(t) $ is maximum — and since $ F'(t) $ is strictly decreasing for $ t > 0 $, maximum at $ t=0 $.", "Therefore, the maximum population growth rate is:", "$$\n\boxed{25}\n$$", "but expressed as a value, not unit — the rate is 25 frogs per year per year.", "However, in ecological literature, growth rate is often reported as a positive number indicating speed of increase, even if decreasing later.", "But strictly, the maximum value of $ F'(t) $ is 25.", "But wait — let's double-check the algebra.", "We had:", "$$\nF'(t) = \frac{400 - 100t^2}{(t^2 + 4)^2}\n$$", "Let $ h(t) = 400 - 100t^2 $, $ d(t) = (t^2 + 4)^2 $", "At $ t = 0 $: $ h = 400 $, $ d = 16 $ → $ h/d = 25 $", "Now, derivative $ F''(t) $: we earlier found a critical point at $ t = 2\sqrt{3} \approx 3.46 $, where $ F'(t) \approx -2.96 $, so indeed a minimum.", "Since $ F'(t) $ starts at 25 and decreases, the global maximum of $ F'(t) $ on $ [0, \infty) $ is $ \boxed{25} $.", "But is this correct? In many scientific contexts, “maximum growth rate” refers to the largest absolute magnitude of increase, but here, since $ F'(t) $ only goes from 25 to 0, the answer is 25.", "However, caution: in population dynamics, “maximum growth rate” often refers to $ \max F'(t) $, which is 25.", "But let’s verify with an alternate method: using AM-GM or substitution.", "Let $ x = t^2 \geq 0 $", "Then:", "$$\nF'(x) = \frac{400 - 100x}{(x + 4)^2}\n$$", "Define $ h(x) = \frac{400 - 100x}{(x + 4)^2} $, $ x \geq 0 $", "Take derivative:", "$$\nh'(x) = \frac{(-100)(x+4)^2 - (400 - 100x) \cdot 2(x+4)}{(x+4)^4}\n= \frac{-100(x+4) - 2(400 - 100x)}{(x+4)^3}\n= \frac{-100x - 400 - 800 + 200x}{(x+4)^3} = \frac{100x - 1200}{(x+4)^3}\n$$", "Set $ h'(x) = 0 $: $ 100x = 1200 $ → $ x = 12 $", "Then $ h(12) = \frac{400 - 1200}{(16)^2} = \frac{-800}{256} = -\frac{25}{8} $", "So maximum at $ x = 0 $: $ h(0) = 400 / 16 = 25 $", "Thus, maximum growth rate is 25", "But this occurs at $ t = 0 $.", "However, in conservation, we may care about the time when growth is fastest, which is at the beginning.", "Alternatively, if interpreted as “the maximum value of the rate of change”, then 25 is correct.", "But in biological modeling, such a high initial growth rate may indicate intense early reproduction, and the peak derivative confirms degradation of that peak.", "Thus, after rigorous analysis, the maximum population growth rate is:", "$$\n\boxed{25}\n$$", "---", "### Conclusion", "The herpetologist models the frog population with $ F(t) = \frac{100t}{t^2 + 4} $. By differentiating and analyzing $ F'(t) $, we found its maximum occurs at $ t = 0 $, yielding a peak growth rate of 25 individuals per year per year. This represents the fastest rate of population increase in the species’ modeled lifecycle. For conservation planning, understanding this peak helps prioritize interventions during early rapid growth phases.", "---", "Keywords: herpetologist, population modeling, frog species, Madagascar, $ F(t) = \frac{100t}{t^2 + 4} $, growth rate, calculus optimization, conservation biology, maximum population growth, differential calculus in ecology", "---", "Updated Answer (Final Boxed):", "$$\n\boxed{25}\n$$"]