Re: partial differentiation

Hi you derivate are good. Maybe if you explain the question you trying to solve I can help more.

Try the Lagrange multiplier?

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Re: partial differentiation

Well, it might seem silly asking for a solution to my problem, but that is what actually would help, so I can compare it with the solutions for other problems and draw some conclusions as to method from there.

As I wrote earlier, my problem is about minimizing

$\displaystyle A(x,h)=2x^2+4xh$

as long as the solution fulfills

$\displaystyle V(x,h)=x^2 \cdot h=4$

It must be mentioned that I haven't had any relevant calculus courses to actually solve the problem.

Re: partial differentiation

I'll guess you don't know about lagrange multiplier and work without them (anyway this is probably faster without it!)

aight in $\displaystyle V(x,h) = h = \frac{4}{(x^2)}$

plug in the $\displaystyle A(x,h) = 2x^2+\frac{4x*4}{x^2}=2x^2+\frac{16}{x^2}=f(x)$

Now we need to find the min of f(x). This you should be able to!

$\displaystyle \frac{d(f(x))}{dx}= 4x +\frac{-32}{x^3}$

if f'(x) = 0

then $\displaystyle x= +- 2^{\frac{3}{4}$

these iare the only point that might be a min or a max. (I'll leave to you to check if it's a min or a max. and its value.)

edit: Typo mistake, I need to slow down when I type stuff..

hope that helped

Re: partial differentiation

Well, that was actually my first approach. What happened was that I got some quite weird results, but since you again propose the same method (which would imply that it might actually work), I will try redoing it, and I will see what I get. I might have made some algebraic screw-up the first time i did it. I'll post, once I see if it works, but I do recognise $\displaystyle 2^\frac 2 3$ from my own solution.

Re: partial differentiation

its 2^(3/4) not 2^(2/3) watch out!

Re: partial differentiation

BTW, err... From my solution:

$\displaystyle A(x,h)$ with $\displaystyle h$ defined and substituted:

$\displaystyle A(x)=2x^2+4x \cdot \frac {4}{x^2} = 2x^2 + \frac {16x} {x^2} = 2x^2 + \frac {16}{x}$

$\displaystyle A'(x)=4x - \frac{16}{x^2}$

Solve the critical point equation:

$\displaystyle A'(x)=4x - \frac{16}{x^2}=0$

$\displaystyle x=2^\frac{2}{3}$

Same result as earlier :/

BTW:

$\displaystyle A(2^\frac{2}{3}, h) \neq A(2^\frac{2}{3}) $ ...

My reason for suspicion. The $\displaystyle h$ has obviously been found by solving the constraint equation for $\displaystyle h$ with $\displaystyle 2^\frac{2}{3}$ substituted for $\displaystyle x$.

Re: partial differentiation

Watch out I made a mistake when I sustitued the h. but the answer you have there seem good. I'm in my class right now. I'll check this out during the break

Re: partial differentiation

To find H put that value on X in V(x,h).

your answer seem very good to me!

still not sure? Let me know I'll try to be more precise.

Re: partial differentiation

I spent some time researching Lagrange multipliers, and I somewhat got the idea of it, but I have been making some sort of mistake when attempting to apply the method, in the sense that I could not solve the three equations, it yielded ... which might imply errors in terms of both algebra and calculus. There is quite a few places, I might have slipped up (Wondering), so if it's not too time consuming (or actually, it is quite time consuming using Lagrange multipliers, it seems), I would appreciate an example of a solution by that method for reasons of comparison.

Re: partial differentiation

We have $\displaystyle \\A(x,h)=2x^2+4xh\\ V(x,h)=x^2h=4$

we start by evaluating the derivate.

$\displaystyle \\V_x =2xh\\V_h=x^2\\A_x=4x+3h\\A_h=4x\\$

Now that we know the derivate we use the langrage multiplier we end up with these equation

$\displaystyle \\A_x=\lambda V_x\Rightarrow 4x+3h = \lambda (2xh)\\A_h=\lambda V_h\Rightarrow 4x = \lambda (x^2)\\V(x,h)=x^2h=4$

You then have to find the value of x,h and $\displaystyle \lambda$. you should end with x =$\displaystyle 2^{\frac{2}{3}}$

thats pretty much the idea of it. Both method work, its up to you to find the "Best" one for each problem. but you should not have different answer.

Re: partial differentiation

That is actually very similar to what I did. I will go to bed soon, but I'll try redoing it by method of Lagrange tomorrow, and we'll see what happens :)

Re: partial differentiation

Aight let me know how it ended!