Solve the problem.
-Find the extreme values of $f ( x , y , z ) = x + 2 y$ subject to $x + y + z = 1$ and $y ^ { 2 } + z ^ { 2 } = 4$.
A) Maximum: $1 + 2 \sqrt { 2 }$ at $( 1 , \sqrt { 2 } , \sqrt { 2 } ) ;$ minimum: $1 - 2 \sqrt { 2 }$ at $( 1 , - \sqrt { 2 } , \sqrt { 2 } )$
B) Maximum: $1 + 2 \sqrt { 2 }$ at $( 1 , \sqrt { 2 } , - \sqrt { 2 } ) ;$ minimum: $1 - 2 \sqrt { 2 }$ at $( 1 , - \sqrt { 2 } , - \sqrt { 2 }$,
C) Maximum: $1 + 2 \sqrt { 2 }$ at $( 1 , \sqrt { 2 } , \sqrt { 2 } )$; minimum: $1 - 2 \sqrt { 2 }$ at $( 1 , - \sqrt { 2 } , - \sqrt { 2 } )$
D) Maximum: $1 + 2 \sqrt { 2 }$ at $( 1 , \sqrt { 2 } , - \sqrt { 2 } ) ;$ minimum: $1 - 2 \sqrt { 2 }$ at $( 1 , - \sqrt { 2 } , \sqrt { 2 } )$

Question

Quizplus · Accepted Answer

The Answer of Solve the problem.
-Find the extreme values of $f ( x , y , z ) = x + 2 y$ subject to $x + y + z = 1$ and $y ^ { 2 } + z ^ { 2 } = 4$.
A) Maximum: $1 + 2 \sqrt { 2 }$ at $( 1 , \sqrt { 2 } , \sqrt { 2 } ) ;$ minimum: $1 - 2 \sqrt { 2 }$ at $( 1 , - \sqrt { 2 } , \sqrt { 2 } )$
B) Maximum: $1 + 2 \sqrt { 2 }$ at $( 1 , \sqrt { 2 } , - \sqrt { 2 } ) ;$ minimum: $1 - 2 \sqrt { 2 }$ at $( 1 , - \sqrt { 2 } , - \sqrt { 2 }$,
C) Maximum: $1 + 2 \sqrt { 2 }$ at $( 1 , \sqrt { 2 } , \sqrt { 2 } )$; minimum: $1 - 2 \sqrt { 2 }$ at $( 1 , - \sqrt { 2 } , - \sqrt { 2 } )$
D) Maximum: $1 + 2 \sqrt { 2 }$ at $( 1 , \sqrt { 2 } , - \sqrt { 2 } ) ;$ minimum: $1 - 2 \sqrt { 2 }$ at $( 1 , - \sqrt { 2 } , \sqrt { 2 } )$

Solve the Problem f(x,y,z)=x+2yf ( x , y , z ) = x + 2 yf(x,y,z)=x+2y

Solve the Problem $f ( x , y , z ) = x + 2 y$