Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find the maximum value of \(H(V)=2 V V^{-1 / 3} /\left(V+V^{-1 / 3}\right)\) for \(V>0\).

Step-by-Step Solution

Answer

The maximum value of the function is \( \frac{\sqrt[4]{27}}{2} \).

Step 1: Simplify the Function

First, simplify the function \( H(V) = \frac{2V^{2/3}}{V + V^{-1/3}} \). Notice that \(V^{-1/3}\) is the reciprocal of \(V^{1/3}\), so the function can be rewritten as \( H(V) = \frac{2V^{2/3}}{V + \frac{1}{V^{1/3}}} \).

Step 2: Use Substitution

Let's simplify the manipulation by substituting \(u = V^{1/3}\), which implies that \(V = u^3\) and \(V^{2/3} = u^2\). Substituting in, we have \( H(u) = \frac{2u^2}{u^3 + \frac{1}{u}} \).

Step 3: Rewrite the Function in Terms of u

Now, rewrite \(H(u)\) as \( H(u) = \frac{2u^3}{u^4 + 1} \). This is an easier form of the function to differentiate.

Step 4: Differentiate to Find Critical Points

Differentiate \( H(u) \) with respect to \( u \) using the quotient rule. Set the derivative \( H'(u) = \frac{d}{du} \left( \frac{2u^3}{u^4 + 1} \right) \) equal to zero to find the critical points.

Step 5: Calculate the Derivative

Apply the quotient rule: \[ H'(u) = \frac{(2u^3)'(u^4 + 1) - 2u^3(u^4 + 1)'}{(u^4 + 1)^2} = \frac{6u^2(u^4 + 1) - 8u^6}{(u^4 + 1)^2} \]. This simplifies to \( \frac{6u^6 + 6u^2 - 8u^6}{(u^4 + 1)^2} = \frac{-2u^6 + 6u^2}{(u^4 + 1)^2} \).

Step 6: Solve the Equation

Set the numerator of \( H'(u) = 0 \), which gives \(-2u^6 + 6u^2 = 0 \). Factor out \(2u^2\), resulting in \( -2u^2(u^4 - 3) = 0 \). This implies \(u^4 = 3\) or \(u = 0\), but since \(u\) must be positive, we have \(u^4 = 3\). Thus, \(u = 3^{1/4}\).

Step 7: Calculate the Maximum Value

Substitute \( u = 3^{1/4} \) back into the simplified function \( H(u) = \frac{2u^3}{u^4 + 1} \). Evaluate it to find \( H(3^{1/4}) = \frac{2 (3^{3/4})}{3 + 1} = \frac{2 \sqrt[4]{27}}{4} \). Simplify this to \( \frac{\sqrt[4]{27}}{2} \).

Step 8: Confirm the Maximum Point

To ensure this is a maximum, check the sign of \( H'(u) \) around \( u = 3^{1/4} \). The derivative changes sign, confirming it as a maximum point.

Key Concepts

Quotient Rule

The Quotient Rule is a method used in calculus to differentiate functions that are written as a ratio of two other functions. Essentially, it gives us a systematic way to handle derivatives of ratios, often simplifying processes that would otherwise be more complex.

This rule states that if you have a function written as a quotient \( \frac{f(x)}{g(x)} \), the derivative is:

• \( \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \)

In our exercise, the function is expressed in the form \( H(u) = \frac{2u^3}{u^4 + 1} \).

Applying the Quotient Rule involves taking the derivative of the numerator \((2u^3)\) and the denominator \((u^4 + 1)\), then substituting these into the formula. This approach is central to finding critical points, which is the next step in optimization.

Critical Points

Critical Points are values of the variable where the derivative of a function is zero or undefined. They are important in calculus for finding maxima, minima, or saddle points of a function.

For the function \( H(u) = \frac{2u^3}{u^4 + 1} \), we calculate the derivative using the Quotient Rule. This results in \( H'(u) = \frac{-2u^6 + 6u^2}{(u^4 + 1)^2} \). To find the critical points, we set the derivative equal to zero:

• \( -2u^6 + 6u^2 = 0 \)

Factoring gives us \( 2u^2(u^4 - 3) = 0 \). Since \( u = 0 \) is not a valid solution (as \( u \) must be positive), we solve \( u^4 = 3 \), yielding \( u = 3^{1/4} \).

Thus, the critical point found is where \( u = 3^{1/4} \), and verifying this through further analysis shows that it represents a maximum for \( H(u) \).

Function Simplification

Function Simplification involves rewriting a complex function in a more manageable form, which often makes derivative calculation and subsequent optimizations feasible.

Initially, our function was given as \( H(V) = \frac{2V^{2/3}}{V + V^{-1/3}} \).

By substituting \( u = V^{1/3} \), we simplify the function to \( H(u) = \frac{2u^2}{u^3 + \frac{1}{u}} \), which then further simplifies to \( H(u) = \frac{2u^3}{u^4 + 1} \). This form is much easier to differentiate and analyze.

Such simplifications are often achieved through clever substitutions that transform the variable and help reveal hidden properties of the function, making the math more straightforward and allowing us to apply rules like the Quotient Rule effectively.