Answer
Maximum:$f(\dfrac{1}{2},\dfrac{1}{2}, \dfrac{1}{2},\dfrac{1}{2}) =2$ and minimum: $f(-\dfrac{1}{2},-\dfrac{1}{2}, -\dfrac{1}{2},-\dfrac{1}{2}) =-2$
Work Step by Step
Our aim is to calculate the extreme values with the help of Lagrange Multipliers Method subject to the given constraints. For this, we have:$\nabla f(x,y)=\lambda \nabla g(x,y)$
This yields $\nabla f=\lt 1,1,1,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z,2t\gt$
From the given constraint condition $x^2+y^2+z^2+t^2=1$ we get, $1=\lambda 2x ,1=\lambda 2y,1=\lambda 2z,1=\lambda 2t$
Simplify. Thus, $x=y=z=t$
Since, $g(x,y)=x^2+y^2+z^2+t^2=1 \implies$ $x=\pm \dfrac{1}{2}$
Also, $x=y=z=t= \pm \dfrac{1}{2}$
Hence, Maximum value is $f(\dfrac{1}{2},\dfrac{1}{2}, \dfrac{1}{2},\dfrac{1}{2}) =2$ and minimum value is $f(-\dfrac{1}{2},-\dfrac{1}{2}, -\dfrac{1}{2},-\dfrac{1}{2}) =-2$
