Implicit Differentiation

[funkdog]

Howdy Folks,

I am wondering if you can give me a hand or point me in the right direction.

So, I have the following equation. I have changed the variables to letters to keep it simple.

X = (B / 2C) + (2D / Y) + (2K / Y)

My task is to max/min this equation for all the different variables. None of these are constants.

For, example. If I want to max/min for C, I think I would differentiate this equation with respect to C, set to 0, and solve.

However, I have only taken a basic calc. course, and am not totally confident on my implicit differentiation -- as it wasn't taught in the class I took.

I understand the concept - but not sure I can totally put it into practice. Most confusing to me is what to do with the (B / 2C) - which has both the term (C) I want differentiate this equation with respect to , and another term B.

The biggest help to me (not sure if anyone has the time) would be to, show me the result for max/min this equation for a few of the variables. For example, if you could show me how the result differs when we differentiate this equation with respect to C versus B or X or D or K or Y.

Whatever help you could give would be greatly appreciated.

Cheers

M

 

[soroban]

Hello, funkdog!

If you're not familiar with Partial Derivatives,
. . I'm not sure you should be working on this problem.

So, I have the following equation. I have changed the variables to letters to keep it simple.

. . \(\displaystyle X \:= \:\frac{B}{2C}\,+\,\frac{2D}{Y}\,+\,\frac{2K}{Y}\)

My task is to max/min this equation for all the different variables. None of these are constants.

For example: if I want to max/min for C, I would differentiate this equation with respect to C,
set to 0, and solve. .I agree.

However, I am not totally confident on my implicit differentiation.
It's not implicit differentiation ... it's partial differentiation.

With partial differentiation, only one of the variables is considered.
. . The others are trated as constant.


Differentiate with respect to \(\displaystyle B\). .We have: \(\displaystyle \L\:X\:=\:\frac{1}{2C}B\,+\,\frac{2D}{y}\,+\,\frac{2K}{y}\)
(Note: only the \(\displaystyle B\) is a variable; \(\displaystyle C,\,K,\,y\) are all constants.)

. . Then: \(\displaystyle \L\:\frac{dX}{dB} \:=\:\frac{1}{2C}\)


Differentiate with respect to \(\displaystyle C\). .We have: \(\displaystyle \L\:X\:=\:\frac{B}{2}C^{-1}\,+\,\frac{2D}{y}\,+\,\frac{2K}{y}\)

. . Then: \(\displaystyle \L\:\frac{dX}{dC} \:=\:-\frac{B}{2}\cdot C^{-2}\)


Differentiate with respect to \(\displaystyle D\). .We have: \(\displaystyle \L\:X\:=\:\frac{B}{2C}\,+\,\frac{2}{y}\cdot D\,+\,\frac{2K}{y}\)

. . Then: \(\displaystyle \L\:\frac{dX}{dD} \:=\:\frac{2}{y}\)


Differentiate with respect to \(\displaystyle K\). .We have: \(\displaystyle \L\:X\:=\:\frac{B}{2C}\,+\,\frac{2D}{y}\,+\,\frac{2}{y}\cdot K\)

. . Then: \(\displaystyle \L\:\frac{dX}{dK} \:=\:\frac{2}{y}\)


Differentiate with respect to \(\displaystyle y\). .We have: \(\displaystyle \L\:X \:=\:\frac{B}{2C}\,+\,2D\cdot y^{-1}\,+\,2K\cdot y^{-2}\)

. . Then: \(\displaystyle \L\:\frac{dX}{dy}\:=\:-2Dy^{-2}\,-\,2Ky^{-2}\:=\:\frac{-2(D\,+\,K)}{y^2}\)

 

[funkdog]

Follow up and thank you so much!

You are a life saver thank you so much! This is really hard to teach myself and I am being asked to do it as part of a project I am working on. Once I see how to do the process once - I think I should be good.

Here, was the actual task I was given:

When you finish all of the above, you may have an estimate of X, but you should think think about the concept of how to minimize X. This would mean writing a general formula for X and then differentiating this formula with respect to the quantitity you want to optimize. For example, you could differentiate with respect to B, and see if you set this to zero, thereby working out a formula for optimum B.

Ok, so here is my final question(s) :wink:

1) Have we done this - by finding the partial derivative? It looks like we have.

2) Now if I set these equal to zero ... aren't I going to end up with zero for some of the answers .. or maybe I just don't understand fully what that means. Could you just finish off a few of them .. so I make sure I understand the process.

Again, thanks so much. If you ever need help with computer programming or other stuff - I'd be there for you.

Cheers

M

 

[soroban]

Hello, funkdog!

This is a strange problem . . . there are no maximums or minimums.
Could the original function be wrong?


We have this function: \(\displaystyle \:X \:= \:\frac{B}{2C}\,+\,\frac{2D}{Y}\,+\,\frac{2K}{Y}\)


Differentiate with respect to \(\displaystyle B:\;\frac{dX}{dB} \:=\:\frac{1}{2C}\)

Differentiate with respect to \(\displaystyle C:\;\frac{dX}{dC} \:=\:-\frac{B}{2}\cdot C^{-2}\:=\:\frac{B}{2C^2}\)

Differentiate with respect to \(\displaystyle D:\;\frac{dX}{dD} \:=\:\frac{2}{Y}\)

Differentiate with respect to \(\displaystyle K:\;\frac{dX}{dK} \:=\:\frac{2}{Y}\)

Differentiate with respect to \(\displaystyle Y:\;\frac{dX}{dY}\:=\:-2DY^{-2}\,-\,2KY^{-2}\:=\:\frac{-2(D\,+\,K)}{Y^2}\)


Now, for example, in the first one, we differentiated with respect to \(\displaystyle B.\)
. . We set that derivative equal to zero and solve for \(\displaystyle B.\)
That value of \(\displaystyle B\) should gives us a maximum or minimum value for \(\displaystyle X.\)

So we have: \(\displaystyle \L\frac{dX}{dB} \:=\:\frac{1}{2C} \:=\:0\)

But there is no value of \(\displaystyle B\) that makes the expression equal zero.

Hence, there is no value of \(\displaystyle B\) which maximizes or minimizes the value of \(\displaystyle X.\)


And we find that this is true for all the derivatives we have.

So there are no maximums or minimum for \(\displaystyle X.\)

 

[funkdog]

I will check

Ok, well first off -- at least I am not crazy.
I was wondering why I couldn't get max or mins or these ...
So, you confirmed that for me - there is none for this equation.

Ok, I may have this original equation wrong. I am going to check that and I will show you what I come up with.

Again, thanks for your help -


M

 

[funkdog]

Thanks for the help.

Soroban,

I really appreciated you taking the time to help me.

I have made a $7 dollar donation to http://schoolswithoutborders.com in your name.

I won't be able to do this every time I need help - but I just wanted to show you how much it meant to me.

I now was able to teach myself all about partial derivatives! And, I was able to determine why this function had no max or mins.


Cheers,

Matt

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