What is the Mistake in Calculating Power Dissipation in a Resistor?

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

Power dissipated in the 2 Ohms resistor will be maximum when current is maximum .This will occur when total resistance is minimum .

If I think logically then I do get the right answer . But on trying to arrive at the result mathematically I think I am making some silly mistake .

Total resistance is Z = (10R+16)/(8+R)

For Z to be minimum , dz/dR = 0

Calculations give me a weird result 64 = 0

What is my mistake ?

 

[.Scott]

Power dissipated in the 2 Ohms resistor will be maximum when current is maximum .This will occur when total resistance is minimum.

Correct.

Total resistance is Z = (10R+16)/(8+R)

Correct.

For Z to be minimum , dz/dR = 0

But there is another restriction on R: R>=0

Calculations give me a weird result 64 = 0

That's because the minimum (and maximum) Z occurs when R=-8 and the denominator goes to zero.
But the problem does not allow for a negative R.

 

[mfb]

Your maximum is not a point where the derivative is zero. It is a point at the edge of the allowed range for R (R cannot be negative).

Here is a plot of the current

 

[Jahnavi]

That's because the minimum (and maximum) Z occurs when R=-8 and the denominator goes to zero.

How did you find that ?

 

[.Scott]

How did you find that ?

You have R+8 in the denominator. Solve for R+8=0.

 

[Jahnavi]

Your maximum is not a point where the derivative is zero.

Are you implying that maximum or minimum can occur either at points where derivative is zero OR at end points of allowed domain ?

 

[Jahnavi]

You have R+8 in the denominator. Solve for R+8=0.

Sorry . But that's not how we find maximum/minimum value of an expression .

 

[.Scott]

Sorry . But that's not how we find maximum/minimum value of an expression .

Well. It's how I found it.
The method you used will work for expressions that do not go to infinity - and which have derivatives.
The restrictions are actually more than that - but I think you want to keep the math at the secondary school level.

 

[mfb]

Are you implying that maximum or minimum can occur either at points where derivative is zero OR at end points of allowed domain ?

Yes. You have an example here.

Another example is the maximum of f(x)=x² in the range of 0 to 2. Clearly the maximum is at x=2 where the derivative is not zero.

 

[Ray Vickson]

Are you implying that maximum or minimum can occur either at points where derivative is zero OR at end points of allowed domain ?

Yes. Look at the simple example of f(x) = x. The minimum of f(x) on 0 ≤ x ≤ 1 is at x = 0, and the maximum on that interval is at x=1. The derivative = 1 at both of those optimal points.

The theorem you are trying to (mis-)use is that if a differentiable function f(x) has a maximum (or a minimum) at some x in the OPEN interval a < x < b, then the derivative vanishes at x. On a CLOSED interval a ≤ x ≤ b, the derivative = 0 an an interior optimum, but not necessarily at an end-point optimum.

 

[Jahnavi]

Yes. You have an example here.

Another example is the maximum of f(x)=x² in the range of 0 to 2. Clearly the maximum is at x=2 where the derivative is not zero.

Thanks !