## Calculus: Early Transcendentals 8th Edition

Maximum:$f(\dfrac{1}{\sqrt n},\dfrac{1}{\sqrt n}, ......,\dfrac{1}{\sqrt n}) =\sqrt n$ and minimum: $f(-\dfrac{1}{\sqrt n},-\dfrac{1}{\sqrt n}, ......,-\dfrac{1}{\sqrt n}) =-\sqrt n$
Use Lagrange Multipliers Method: $\nabla f(x_1,x_2,...,x_n)=\lambda \nabla g(x_1,x_2,....,x_n)$ This yields $\nabla f=\lt 1,1,1,....,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x_1,2x_2,2x_3,....,2x_n\gt$ Using the constraint condition $x^2+y^2+z^2+t^2=1$ we get, $1=\lambda 2x_1 ,...,1=\lambda 2x_n$ After solving, we get $x_1=x_2=x_3=...=N$ Thus, $x_1=x_2=....= \pm \dfrac{1}{\sqrt n}$ Hence, Maximum:$f(\dfrac{1}{\sqrt n},\dfrac{1}{\sqrt n}, ......,\dfrac{1}{\sqrt n}) =\sqrt n$ and minimum: $f(-\dfrac{1}{\sqrt n},-\dfrac{1}{\sqrt n}, ......,-\dfrac{1}{\sqrt n}) =-\sqrt n$