Answer
Maximum:$f(\dfrac{1}{\sqrt n},\dfrac{1}{\sqrt n}, ......,\dfrac{1}{\sqrt n}) =\sqrt n$ and minimum: $f(-\dfrac{1}{\sqrt n},-\dfrac{1}{\sqrt n}, ......,-\dfrac{1}{\sqrt n}) =-\sqrt n$
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:$\nabla f(x,y)=\lambda \nabla g(x,y)$
Re-write as: $\nabla f(x_1,x_2,...,x_n)=\lambda \nabla g(x_1,x_2,....,x_n)$
This yields $\nabla f=\lt 1,1,1,....,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x_1,2x_2,2x_3,....,2x_n\gt$
As per given constraint condition, we have $x^2+y^2+z^2+t^2=1$ we get, $1=\lambda 2x_1 ,...,1=\lambda 2x_n$
Simplify. Thus, $x_1=x_2=x_3=...=N$
Now, $x_1=x_2=....= \pm \dfrac{1}{\sqrt n}$
We find that the Maximum value is $f(\dfrac{1}{\sqrt n},\dfrac{1}{\sqrt n}, ......,\dfrac{1}{\sqrt n}) =\sqrt n$ and minimum value is $f(-\dfrac{1}{\sqrt n},-\dfrac{1}{\sqrt n}, ......,-\dfrac{1}{\sqrt n}) =-\sqrt n$
