## Calculus: Early Transcendentals 8th Edition

Nearest $f(\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2})$ and farthest: $f(-1,-1,2)$
Our aim is to calculate the extreme values with the help of the Lagrange Multipliers Method subject to the given constraints. For this, we have:$\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ This yields $\nabla f(x,y,z)=\lt 2x,2y,2z \gt$ and $\lambda_1 g(x,y)= \lt 1,1,2 \gt$ , $\lambda_1 h(x,y,z)= \lt 2x,2y,-1 \gt$ Using the constraint condition $x^2+y^2 =z$, we get, $z=-\dfrac{1}{2}$ Let us consider $x=y$, after using the given conditions, we have $x+z=1$ and $2x^2=z$ Thus, $2x^2+x-1=0 \implies x= \dfrac{1}{2} ,-1$ We conclude that $f(\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2})$ is the point on the ellipse nearest to the origin and $f(-1,-1,2)$ is the point on the ellipse farthest from the origin. Hence, the desired results are: Nearest $f(\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2})$ and farthest: $f(-1,-1,2)$