Question

find maximum: f(x,y,z)=x+2y-2z constraint is x^2+y^2+z^2=4

Answer 1

looks like a standard Lagrange multiplier problem

Answer 2

Remember that the idea of Lagrange multipliers is that if we consider two functions \(f(x,y,z)=x+2y-2z\) and \(g(x,y,z)=x^2+y^2+z^2\) (\(g\) is our new name for the constraint), there will exist Lagrange Multiplier values \(\lambda \) such that \(\vec \nabla f = \lambda \vec \nabla g\) (we will later find the appropriate value for our constraing \(g=4\) later in the problem). This has to do with perpendicularity of level curves and gradient vectors -- I'm sure if you've been taught this you're at least familiar with the theory. Anyway, given that gradient is generically given by \(\vec \nabla A = \left\langle \frac{\partial A}{\partial x}, \frac{\partial A}{\partial y}, \frac{\partial A}{\partial z} \right\rangle\), we can evaluate \(\vec \nabla f = \lambda \vec \nabla g\) for this case. \[\left\langle 1, 2, -2\right\rangle =\lambda \left\langle 2x, 2y, 2z \right\rangle\]We can rewrite this as three simultaneous equations, one per component.\[\left.\begin{align}1&=\lambda 2x \\ 2&=\lambda 2y \\ -2&=\lambda 2z\end{align}\right\} \Rightarrow \left\{\begin{align}x&=\frac{1}{2\lambda}\\ y&=\frac{1}{\lambda} \\ z&=-\frac{1}{\lambda}\end{align}\right. \]Plugging these found values into \(g\) gives us the following.\[g(\lambda)=\left(\frac{1}{2\lambda}\right)^2+\left(\frac{1}{\lambda}\right)^2+\left(-\frac{1}{\lambda}\right)^2=\frac{9}{4\lambda^2}\]Since we know \(g=4\) as our given constraint, we can solve for \(\lambda\).\[g(\lambda)=\frac{9}{4\lambda^2}=4 \Rightarrow \lambda = \pm\frac{4}{3}\]We can use these values to find the associated \((x,y,z)\) coordinates. \[\left(\frac 3 8, \frac 3 4, -\frac 3 4\right) \vee \left(-\frac 3 8, -\frac 3 4, \frac 3 4\right)\]We will plug in these two values into the original equation to see which one is truly the maximum.\[f\left(\frac 3 8, \frac 3 4, -\frac 3 4\right) = \frac 38 + 2\left(\frac 34 \right) - 2\left(-\frac 34\right) = \boxed{\displaystyle \frac{27}{8}}\]\[f\left(-\frac 3 8, -\frac 3 4, \frac 3 4\right) = -\frac 38 + 2\left(-\frac 34 \right) - 2\left(\frac 34\right) = -\frac{27}{8}\]

Answer 3

Whoops! Silly algebraic error! I should have said \(\lambda = \pm\dfrac{3}{4}\), but you get the idea! When you use that value, you'll find the actual maximum abiding by that constrint is \(\boxed{6}\). If you still need help, give me a holler!