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
Use Lagrange Multipliers Method:
$\nabla f(x,y,z,t)=\lambda \nabla g(x,y,z,t)$
This yields $\nabla f=\lt 1,1,1,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z,2t\gt$
Using the constraint condition $x^2+y^2+z^2+t^2=1$ we get, $1=\lambda 2x ,1=\lambda 2y,1=\lambda 2z,1=\lambda 2t$
After solving, we get $x=y=z=t$
Since, $g(x,y)=x^2+y^2+z^2+t^2=1$ gives $x=\pm \dfrac{1}{2}$
Thus, $x=y=z=t= \pm \dfrac{1}{2}$
Hence, Maximum:$f(\dfrac{1}{2},\dfrac{1}{2}, \dfrac{1}{2},\dfrac{1}{2}) =2$ and minimum: $f(-\dfrac{1}{2},-\dfrac{1}{2}, -\dfrac{1}{2},-\dfrac{1}{2}) =-2$
