Problem 30 - Applied Calculus

Question

What value of \(w\) minimizes \(S\) if \(S-5 p w=3 q w^{2}-6 p q\) and \(p\) and \(q\) are positive constants?

Step-by-Step Solution

Answer

The value of \(w\) that minimizes \(S\) is \(w = -\frac{5p}{6q}\).

Step 1: Identify the Objective Function

The goal is to find the value of \(w\) that minimizes \(S\). From the equation \(S - 5pw = 3qw^2 - 6pq\), we identify that \(S = 3qw^2 - 6pq + 5pw\). This is our objective function to minimize.

Step 2: Express the Function in Standard Form

The function \(S = 3qw^2 - 6pq + 5pw\) can be rewritten as \(S(w) = 3qw^2 + 5pw - 6pq\). This is a quadratic function in terms of \(w\).

Step 3: Find the Derivative of the Function

To find the minimum value, we first find the derivative of \(S(w)\) with respect to \(w\). The derivative is \(S'(w) = \frac{d}{dw}(3qw^2 + 5pw - 6pq) = 6qw + 5p\).

Step 4: Determine the Critical Points

Set the derivative equal to zero to find the critical points: \(6qw + 5p = 0\). Solve for \(w\): \(6qw = -5p\), giving \(w = -\frac{5p}{6q}\).

Step 5: Verify it is a Minimum

To ensure that \(w = -\frac{5p}{6q}\) is indeed a minimum, check the second derivative: \(S''(w) = \frac{d^2}{dw^2}(3qw^2 + 5pw) = 6q\). Since \(q > 0\), \(S''(w) = 6q > 0\), confirming that the critical point is a minimum.

Key Concepts

Objective Function

In optimization problems, the objective function is the function that we aim to minimize or maximize. Here, our goal is to minimize the value of the function \(S\). The original equation provided is \(S - 5pw = 3qw^2 - 6pq\). From this, we can express the objective function as \(S(w) = 3qw^2 + 5pw - 6pq\). This equation represents a quadratic function in terms of \(w\).
Remember, the objective function is the foundation of any optimization problem. It's our primary focus, and understanding it is crucial in finding optimum solutions.

Derivative of a Function

Finding the derivative of a function is a core step in calculus used to understand how the function behaves. The derivative helps determine the rates of change and identifies critical points that could be potential minima or maxima.
For our objective function \(S(w) = 3qw^2 + 5pw - 6pq\), the derivative \(S'(w)\) is calculated as follows:

• Differentiate with respect to \(w: S'(w) = \frac{d}{dw}(3qw^2 + 5pw - 6pq) = 6qw + 5p\).

By determining \(S'(w)\), we can analyze where the function's rate of change equals zero, indicating potential critical points.

Critical Points

Critical points occur where the derivative of a function is equal to zero or undefined, signaling potential locations for minima, maxima, or saddle points. By setting \(S'(w) = 6qw + 5p = 0\), we can solve for the critical point of our function:

• Solve for \(w\): \(6qw = -5p\).
• This gives us \(w = -\frac{5p}{6q}\).

Critical points are essential in determining where to test for minima or maxima using further tests like the second derivative test.

Second Derivative Test

The second derivative test is a method used to classify critical points without having to construct a complete graph of the function. It determines whether a given critical point is a minimum or maximum.
For our function \(S(w)\), the second derivative is \(S''(w) = \frac{d^2}{dw^2}(3qw^2 + 5pw - 6pq) = 6q\). Because \(q\) is positive, \(S''(w) = 6q > 0\).
A positive second derivative at the critical point \(w = -\frac{5p}{6q}\) indicates that the function \(S(w)\) is convex at this point, confirming that it is indeed a minimum. Thus, the second derivative test provides reassurance that \(w = -\frac{5p}{6q}\) minimizes our objective function \(S\).