Recent content by Hart

.. oh wait, for say the first one do I get: f_{1}(x,y,z) = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} f_{2}(x,y,z) = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} f_{3}(x,y,z) = \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} f_{4}(x,y,z)...

So as an answer to the first part of the question, I get that the extrema are: f(x,y,z) = \left(\frac{-2\lambda}{3} , \frac{-2\lambda}{3} , \frac{-2\lambda}{3}\right) since \lambda = \frac{\sqrt{3}}{2} .. is this sufficient to state just this? I thought I'd get a list of points maybe...

.. did you post a reply? I got a notification email but it's not showing up here :confused:

OK, so I now know the values of x , y , z and of \lambda which should be: \lambda = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} SO that answers part i of the question (where x, y, and z are all non-zero), yep? Now for part ii (where x = 0), I just do the same but with x = 0? I.e...

.. so from the first three equations I get: x = \frac{-2\lambda}{3} y = \frac{-2\lambda}{3} z = \frac{-2\lambda}{3} .. then square these values and input into the fourth equation, which gives: \frac{4\lambda^{2}}{9} + \frac{4\lambda^{2}}{9} + \frac{4\lambda^{2}}{9} - 1 = 0...

so: 3x + 2\lambda = 0 \implies x = \frac{-2\lambda}{3} ?

I can only see how to solve those first three equations in terms of a quadratic? I.e. that: \frac{\partial F}{\partial x} = 3x^{2} + 2x \lambda = 0 \implies x = \frac{-2\lambda \pm \sqrt{(2\lambda)^{2}}}{6} \implies x = -3\lambda \pm 2\lambda .. is this the right way to go?

\frac{\partial F}{\partial x} = 3x^{2} + 2x \lambda = 0 \frac{\partial F}{\partial y} = 3y^{2} + 2y \lambda = 0 \frac{\partial F}{\partial z} = 3z^{2} + 2z \lambda = 0 \frac{\partial F}{\partial \lambda} = x^{2} + y^{2} + z^{2} - 1 = 0

Ah ok, so it should be: \frac{\partial F}{\partial \lambda} = x^{2} + y^{2} + z^{2} - 1 = 0 yes? .. and then how to I go about getting values of x, y, and \lambda? Not sure how to equate those 4 simulataneous equations :confused:

Ok, good. Yes, I just missed that, so I should have: \frac{\partial F}{\partial x} = 3x^{2} + y^{3} + z^{3} + \lambda \left(2x + y^{2} + z^{2} - 1\right) = 0 \frac{\partial F}{\partial y} = 3y^{2} + x^{3} + z^{3} + \lambda \left(2y + x^{2} + z^{2} - 1\right) = 0 \frac{\partial...

Ermm.. so need to do this: f(x,y) = x^{3} + y^{3} + z^{3} g(x,y) = x^{2} + y^{2} + z^{2} = 1 Hence need to then extremise: F(x,y,\lambda) = f + \lambda y = x^{3} + y^{3} + z^{3} + \lambda \left(x^{2} + y^{2} + z^{2} - 1\right) So need to then calculate partial derivatives of this...

Thought I may have made a mistake with calculations of the partial derivatives :frown: So, should be then: \frac{\partial F}{\partial x} = 2 + 2x\lambda = 0 \frac{\partial F}{\partial x} = 2 + 2y\lambda = 0 \frac{\partial F}{\partial x} = \left(x^{2} + 4y^{2} - 1\right) = 0 ...

.. I was doing in the \lambda way (um.. lagrange multiplier?!) as this is how I was taught a similar problem and hence should really use this way. Looking back through my notes on this, once the values of x and y have been found (as they have been) then just put these into the original...

Homework Statement Considering the surface of a sphere of radius 1 with its centre at coordinates (0,0,0). For the function: f(x,y,z) = x^{3} + y^{3} + z^{3} Need to find the following: (i) All the extrema on the surface which have x, y and z all non-zero simultaneously...

OH.. I just forgot them! :redface: ok, so: f(x,y) = xy - \left(\frac{y^{3}}{3}\right) \frac{\partial f(x,y)}{\partial xy} = 1 - y^{2} \frac{\partial f(x,y)}{\partial xy} = 2y .. correct? (I'm not too good on partial derivations :frown:) So then: 2y = 2\left(-1 -...