Simple minimum/maximum problem

  • Thread starter John O' Meara
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Then I get R^2=a^2+b^2. That's R, so R= \sqrt{a^2+b^2}. So far we have the maximum value of R. Now, for the minimum value of R, we need to find the value of alpha that minimizes R. To do this, we can take the derivative of R with respect to alpha and set it equal to 0. We get R'=sin(alpha)*a-cos(alpha)*b=0. Solving for alpha, we get tan(alpha)=a/b. So, for the minimum value of R, we have R=\sqrt{a^2+b^2}, with alpha=arctan(a/b). Therefore, in summary,
  • #1
John O' Meara
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The voltage E applied to the sending end of a high-pressure transmitting line is connected to the voltage e at the receiving end by the equation [tex] E^2 = ( e cos \theta + a)^2 + ( e sin\theta +b)^2\\ [/tex], where a and b are constants.
Expand the right-hand side of this equation and by expressing [tex] a cos\theta + b sin\theta \\[/tex] in the form [tex] R cos(\theta + \alpha) \\[/tex] show that the maximum and minimum values of R as [tex] \theta [/tex] varies , are e +/- [tex] \sqrt{a^2 +b^2} \\ [/tex]. ( obtaining these results by differentation is a much more difficult method and we would remind our readers that the calculus is not always the best way of solving a problem.
Expanding this I get: [tex] E^2= e^2 cos^2\theta +a^2 +2ae cos\theta + e^2 sin^2\theta + 2eb sin\theta +b^2 \\[/tex], which gives [tex] E^2 = e^2 +2e(a cos\theta + b sin\theta ) +a^2 + b^2 \\[/tex], which can be expressed as
[tex]e^2 + 2e(a cos\theta + b cos(\frac{\pi}{2} - \theta) ) + a^2 + b^2[/tex]. The thing is I do not know how to express R in terms of the coefficients a and b? Thanks for the help.
 
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  • #2
You didn't do what you were asked to do.

You rewrote [itex]\sin\theta[/itex] as [itex]\cos(\pi/2-\theta)[/itex]. That is not what the problem says to do. The problem specifically says to write [itex]a\cos\theta+b\sin\theta[/itex] in the form [itex]R\cos(\theta+\alpha)[/tex]. Simply expand this latter form, equate to [itex]a\cos\theta+b\sin\theta[/itex], and solve for [itex]R[/itex] and [itex]\alpha[/itex].
 
  • #4
Expanding R [tex] cos(\theta + \alpha) \\[/tex] we get[tex] R( \cos\theta \sin\alpha - \sin\theta \sin\alpha) \\ [/tex]. Now [tex] R \cos\theta\cos\alpha - R \sin\theta\sin\alpha = a \cos\theta + b \sin\theta \\ [/tex] When [tex] \theta = 0 R = \frac{a}{\cos\alpha} \\[/tex] When [tex] \theta = \frac{\pi}{2} R = \frac{-b}{\sin\alpha}\\ [/tex] Therefore [tex] \tan\alpha = \frac{-b}{a} [/tex]. I would appreciate help in getting R.
 
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  • #5
Ok, so a=R*cos(alpha) and b=-R*sin(alpha). Square and add both of those equations and use everybodies favorite trig identity.
 

FAQ: Simple minimum/maximum problem

What is a simple minimum/maximum problem?

A simple minimum/maximum problem is a type of optimization problem where the goal is to find the minimum or maximum value of a function, subject to certain constraints. The function can have one or more variables, and the constraints can be equations or inequalities.

How do you solve a simple minimum/maximum problem?

To solve a simple minimum/maximum problem, you need to first identify the function and the constraints. Then, you can use various mathematical techniques such as differentiation or linear programming to find the optimal solution. It is also important to check for any critical points or boundary points that may affect the solution.

What are some real-life applications of simple minimum/maximum problems?

Simple minimum/maximum problems are commonly used in engineering, economics, and other fields where optimization is important. For example, they can be used to determine the most cost-effective production levels for a company, the optimal design for a bridge, or the best investment portfolio.

How can I ensure that my solution to a simple minimum/maximum problem is correct?

To ensure the accuracy of your solution, it is important to double-check your calculations and make sure you have considered all possible constraints. You can also use a graphing calculator or software to visualize the function and constraints and verify your solution.

Are there any limitations to simple minimum/maximum problems?

While simple minimum/maximum problems can be a powerful tool for optimization, they may not always provide the most practical or realistic solutions. This is because they assume that the function and constraints are linear and do not take into account external factors or uncertainties. It is important to carefully consider the assumptions and limitations of the problem before using this approach.

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