Differentiation maximum problem

In summary, the power of an electrical circuit can be calculated using the formula P=((E^2)R)/(R + r)^2, where E is the emf of the battery. To find the maximum power, the formula can be differentiated with respect to R, resulting in the expression E^2*(r-R)/(R+r)^3. This is equal to 0 when r=R, indicating the maximum power is achieved when the resistance of the circuit matches the internal resistance of the battery.
  • #1
Joza
139
0
Power of an electrical circuit is equal to current squared times resistance.

My expression for current is: E/(R + r), where E is emf of battery.

So, my expression for Power, P, is:

P=((E^2)R)/(R + r)^2

I would like to differentiate P with respect to R, to find where the slope is zero, ie., the maximum power.
I am having some trouble though. I used the quotient rule and got:

((R + r)E^2 - 2RE^2)/(R + r)^3

Is this correct? I need to know where this is zero, but I am a bit unsure about where to go from here.
 
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  • #2
Yes, that is correct. But I think you will make the problem much easier by combining those "R2"s!
[tex]\frac{(R+r)E^2- 2RE^2}{(R+r)^3}= E^2\frac{r- R}{(R+r)^3}[/itex]
and, of course, a fraction is only 0 where the numerator is 0.
 
  • #3
Ah, excellent!

I guess I was a bit too tired to realize I should expand it.

And of course it's zero then...:tongue: why didn't I see that!

That agrees perfectly, because it's 0 when r=R. And that is what I was looking for!

Brilliant mate, cheers!:biggrin:
 

1. What is a differentiation maximum problem?

A differentiation maximum problem is a type of optimization problem that involves finding the maximum value of a function by using the derivative, or rate of change, of the function. In other words, it is a way to find the highest point on a curve or surface.

2. How is differentiation used to solve a maximum problem?

Differentiation is used to solve a maximum problem by finding the derivative of the function and setting it equal to zero. This gives us the critical points, or points where the function may have a maximum value. We then use the second derivative test to determine if the critical points are maximum points.

3. What is the second derivative test?

The second derivative test is a method used to determine if a critical point is a maximum, minimum, or neither. It involves taking the second derivative of the function and plugging in the critical point. If the second derivative is positive, the critical point is a minimum. If the second derivative is negative, the critical point is a maximum. If the second derivative is zero, the test is inconclusive and further analysis is needed.

4. Are there any real-world applications of differentiation maximum problems?

Yes, differentiation maximum problems have many real-world applications. For example, they can be used to find the maximum profit or minimum cost in a business, the optimal production level in manufacturing, or the maximum velocity of a falling object due to gravity. They are also used in fields such as economics, physics, and engineering.

5. What are some strategies for solving differentiation maximum problems?

There are a few strategies for solving differentiation maximum problems, including using the first and second derivative tests, setting up and solving optimization equations, and graphing the function to visually identify the maximum point. It is also helpful to understand the properties of different types of functions, such as polynomial, exponential, and trigonometric functions, to better approach the problem. Practice and familiarity with these strategies can also improve problem-solving skills.

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