Problem 33 - 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: Rearrange equation

Rearrange the given equation for \(S\):\[S = 3qw^2 + 5pw - 6pq\]Now, we have an equation for \(S\) in terms of \(w\).

Step 2: Identify the function to minimize

Recognize that under standard mathematical operations, we need to minimize the quadratic function in terms of \(w\):\[S(w) = 3qw^2 + 5pw - 6pq\]This is a quadratic function where the coefficient of \(w^2\) is positive, indicating a parabola that opens upwards.

Step 3: Find the critical point

Differentiate \(S(w)\) with respect to \(w\) to find the critical point:\[\frac{dS}{dw} = 6qw + 5p\]Set the derivative equal to zero to find the value of \(w\) that gives a minimum:\[6qw + 5p = 0\]

Step 4: Solve for \(w\)

Solve for \(w\) in the equation derived from differentiating:\[w = \frac{-5p}{6q}\]This is the value of \(w\) that minimizes \(S\).

Step 5: Verify the second derivative

Check if the second derivative confirms a minimum:\[\frac{d^2S}{dw^2} = 6q\]Since \(q > 0\), \(6q > 0\), confirming that the point is indeed a minimum for \(S\).

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial that has a degree of two. In general, it is expressed in the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). This ensures the function graphs as a parabola. The quadratic function provided in the problem is \(S(w) = 3qw^2 + 5pw - 6pq\), where \(3q\) is the coefficient of \(w^2\).

In this expression:

• \(3q\) is the leading coefficient, and it determines how wide or narrow the parabola opens.
• \(5p\) is the linear coefficient, influencing the vertex's location horizontally.
• \(-6pq\) is the constant term, which affects the vertical placement of the parabola on the graph.

Knowing that the parabola opens upwards (as the leading coefficient \(3q\) is positive), we can determine that a minimum point exists.

Differentiation

Differentiation is a process in calculus used to find the rate at which a function is changing at any given point. In optimization problems, differentiation helps in finding where a function reaches its maximum or minimum values. To differentiate the quadratic function \(S(w) = 3qw^2 + 5pw - 6pq\) with respect to \(w\), we apply the power rule.

Here are the steps for differentiation:

• The derivative of \(3qw^2\) becomes \(6qw\) by using the power rule, bringing down the exponent \(2\) and multiplying by \(3q\).
• The derivative of \(5pw\) with respect to \(w\) is simply \(5p\), as \(p\) is a constant.
• The constant \(-6pq\) vanishes as its derivative is zero.

This yields the derivative \(\frac{dS}{dw} = 6qw + 5p\). Differentiating allows us to find critical points where the function could have potential minima or maxima.

Critical Points

Critical points are values in the domain of a function where the derivative is zero or undefined, which could indicate the function's local minima, maxima, or saddle points. For the quadratic function in the problem, the critical point is found by setting its derivative: \(\frac{dS}{dw} = 6qw + 5p\) equal to zero.

Solving the equation:

• Set \(6qw + 5p = 0\).
• Solve for \(w\), resulting in \(w = \frac{-5p}{6q}\).

This value of \(w\) represents the point where the function changes its rate from decreasing to increasing, confirming it as a local minimum given the nature of quadratic functions.

Second Derivative Test

The second derivative test is a method used to determine the concavity of a function and confirm whether a critical point is a minimum or a maximum. For this purpose, we will examine the second derivative of the function:

The second derivative of \(S(w) = 3qw^2 + 5pw - 6pq\) is:

• Differentiate the first derivative \(\frac{dS}{dw} = 6qw + 5p\), resulting in \(\frac{d^2S}{dw^2} = 6q\).

Here, since \(q\) is positive by problem condition, \(6q\) is also positive.

A positive second derivative indicates that the function is concave upwards at this critical point, meaning that the point \(w = \frac{-5p}{6q}\) is indeed a local minimum. This confirms that the value we calculated is the minimum point of the quadratic function.