Find the highest and lowest points made by the union of these two functions using Lagrange Multipliers.
$x^2+y^2+z^2 = 16$
$(x+1)^2+(y+1)^2+(z+1)^2 = 27$
I got the basics down, I used the first function as my constraint.
$2(x+1) = \lambda\times2x$
$2(y+1) = \lambda\times 2y$
$2(z+1) = \lambda\times2z$
$x^2+y^2+z^2 = 16$
and after solving I think it's all $x=y=z=\pm\dfrac{4}{\sqrt{3}}$
I just don't know how to interpret these results.
What does this mean? and how do I use this to answer
my problem?
Thank you for your time.
I think you are missing something. As @copper.hat noted you don't have nothing to optimize. I think that the two equations you uploaded are constrains, but the function you want to maximize/minimize isn't posted.
If that's not the case, add those two and you'll get that both minima and the maxima of the function is $27 + 16 = 43$.
If you want to find a solution for the system of equations, just subtract the first equation from the second and we have:
$$x + y + z = 4$$ $$x^2 + y^2 + z^2 = 16$$
Then there are infinite amount of solution. Set $x^3 + y^3 + z^3$ to any value you like and then using Newton's Identities you'll be able to obtain a cubic polynomial, where $x,y,z$ will be roots of that polynomial.
And from the thing you've posted I think you want to maximize/minimize the function
$$f(x,y,z) = (x+1)^2 + (y+1)^2 + (z+1)^2$$
under constrain:
$$g(x,y,z) = x^2 + y^2 + z^2 - 16 = 0$$
To make things even simplier expand the first function and you'll have:
$$f(x,y,z) = (x^2 + y^2 + z^2) + 2(x+y+z) + 3 = 2(x+y+z) + 19$$
Now apply Lagrange multipliers:
$$F(x,y,z,\lambda) = 2(x+y+z) + 19 - \lambda(x^2 + y^2 + z^2 - 16)$$
Now take partial derivatives and set them to 0.
$$F_x = 2 - 2x\lambda = 0$$ $$F_y = 2 - 2y\lambda = 0$$ $$F_z = 2 - 2z\lambda = 0$$ $$F_{\lambda} = x^2 + y^2 + z^2 - 16 = 0$$
Substituting $x = y = z \frac{1}{\lambda}$ from the first partial derivatives into the constarin we have:
$$\frac{3}{\lambda^2} = 16$$ $$\lambda = \pm \frac{\sqrt{3}}{4}$$
Substitute back and get that $ x = y = z = \pm \frac{4}{\sqrt{3}}$
Plug the values for $x,y,z$ into the initial function and you'll get that the maxima occurs at $f(\frac{\sqrt{3}}{4},\frac{\sqrt{3}}{4},\frac{\sqrt{3}}{4}) = 19 + 8\sqrt{3}$ and the minima at $f(\frac{- \sqrt{3}}{4}, - \frac{\sqrt{3}}{4}, - \frac{\sqrt{3}}{4}) = 19 - 8\sqrt{3}$