# Bah calculus calculus & more calculus

1. May 13, 2010

### mathstruggle

got a simple problem.
the question is find min value for f=x*y+(z^2) with constraints 2*x -y=8, and co-ordinates where it occurs.
so far what i did.
▽f=λ*▽g
F(x,y,z)=f(x,y)-λg(x,y,z)
F=(xy+z^2)-λ(2x-y-8)

y=2λ
x=λ

i got λ=-2

is this right? and is that the only λ value?

how do i even start to find equilibrium solutions & general solution to ((t/2)-2)*sin(y), we weren't taught or shown how to find it involving sin,cos, tan?
for general solution i should take it as separable to find the general solution? but how do i start? sin, cos, tan bah

Last edited: May 13, 2010
2. May 13, 2010

### lanedance

you could probaby substitute directly into that function & minimise

3. May 13, 2010

### lanedance

otherwise lagrange multipliers are alway good...

4. May 13, 2010

### HallsofIvy

I think you are trying to use "Lagrange multipliers" as lanedance suggested but you seem to have no idea how to do that.

You write, correctly, that $\nabla f= \lambda \nabla g$ but then "F(x,y,z)=f(x,y)-λg(x,y,z)" and "F=(xy+z^2)-λ(2x-y-8)" which have nothing to do with what you wrote previously.

$\nabla f$ is the vector $y\vec{i}+ x\vec{j}+ 2z\vec{k}$ and $\nabla g= 2\vec{i}- \vec{j}$.

"$\nabla f= \lambda \nabla g$" is now
$y\vec{i}+ x\vec{j}+ 2z\vec{k}= \lambda 2\vec{i}- \vec{j}$

Looking at the individual components of that, $y= 2\lambda$, $x= -\lambda$, and $z= 0$.

Now, you have $y= 2\lambda$ but you have $x= \lambda[\itex] rather than [itex]x= -\lambda$. Perhaps that is just a typo. In any case, they do not give "$\lambda= -2$ because you have no reason to believe x= -2 or y= -4!.

Rather, $x= -\lambda$ says that $\lambda= -x$ and so $y= 2\lambda= -2x$. Putting that into the constraint 2x- y= 8 gives 2x+ 2x= 4x= 8 so x= 2 and y= -4. The solution is (2, -4, 0).

Is this a completely different problem? Then what is the problem? I associate "equilibrium solutions & general solution" with differential equations but you give no differential equation. Are you talking about dy/dt= ((t/2)- 2)sin(y) or some other problem? Please state the entire problem.

And a general suggestion: if you want people who are really good at math to help you (hopefully more politely than I did), stop dissing mathematics!

5. May 19, 2010

### mathstruggle

i didn't diss maths. why would i diss maths if im majoring in maths? its called sense of humour. last time im using this forum, its like dictatorship here no freedom.