Answer
$(\dfrac{100}{3} ,\dfrac{100}{3} ,\dfrac{100}{3})$
Work Step by Step
Given: $x+y+z=100$
Re-write the given equations as: $z=100-x-y$
To solve this problem we will take the help of Second derivative test that suggests the following conditions to determine the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
i) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
ii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
iii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
$D(\dfrac{100}{3},\dfrac{100}{3})=\dfrac{200}{3} \gt 0$ and $f_{xx}=-\dfrac{200}{3} \lt 0$
Hence, our result is:
$(\dfrac{100}{3} ,\dfrac{100}{3} ,\dfrac{100}{3})$
