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Chapter 4: Problem 33

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?

Short Answer

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

Step by step solution

01

Rearrange equation

Rearrange the given equation for \(S\):\[S = 3qw^2 + 5pw - 6pq\]Now, we have an equation for \(S\) in terms of \(w\).
02

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.
03

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:\[6qw + 5p = 0\]
04

Solve for \(w\)

Solve for \(w\) in the equation derived from differentiating:\[w = \frac{-5p}{6q}\]This is the value of \(w\) that minimizes \(S\).
05

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 \(S\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

If \(t\) is in years since \(1990,\) one model for the population of the world, \(P,\) in billions, is $$P=\frac{40}{1+11 e^{-0.08 t}}$$ (a) What does this model predict for the maximum sustainable population of the world? (b) Graph \(P\) against \(t\). (c) According to this model, when will the earth's population reach 20 billion? 39.9 billion?

A business sells an item at a constant rate of \(r\) units per month. It reorders in batches of \(q\) units, at a cost of \(a+b q\) dollars per order. Storage costs are \(k\) dollars per item per month, and, on average, \(q / 2\) items are in storage, waiting to be sold. [Assume \(r, a, b, k\) are positive constants.] (a) How often does the business reorder? (b) What is the average monthly cost of reordering? (c) What is the total monthly cost, \(C\) of ordering and storage? (d) Obtain Wilson's lot size formula, the optimal batch size which minimizes cost.

(a) A cruise line offers a trip for \(\$ 2000\) per passenger. If at least 100 passengers sign up, the price is reduced for all the passengers by \(\$ 10\) for every additional passenger (beyond 100 ) who goes on the trip. The boat can accommodate 250 passengers. What number of passengers maximizes the cruise line's total revenue? What price does each passenger pay then? (b) The cost to the cruise line for \(n\) passengers is \(80,000+400 n .\) What is the maximum profit that the cruise line can make on one trip? How many passengers must sign up for the maximum to be reached and what price will each pay?

Let \(C(q)\) represent the cost, \(R(q)\) the revenue, and \(\pi(q)\) the total profit, in dollars, of producing \(q\) items. (a) If \(C^{\prime}(50)=75\) and \(R^{\prime}(50)=84,\) approximately how much profit is earned by the \(51^{\text {st }}\) item? (b) If \(C^{\prime}(90)=71\) and \(R^{\prime}(90)=68,\) approximately how much profit is earned by the \(91^{\text {st }}\) item? (c) If \(\pi(q)\) is a maximum when \(q=78,\) how do you think \(C^{\prime}(78)\) and \(R^{\prime}(78)\) compare? Explain.

Would you expect the demand for high-definition television sets to be elastic or inelastic? Explain.
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