Question

The extrema value for the function f(x,y)= x^2 + y^2 under the condition x+y=1, is
1/2
O -112
O 14
-1/4​

Answer 1

Answer:

Step-by-step explanation:

f(x, y) − f(a, b) > 0,

for all (x, y) 6= (a, b) in the domain of f then we say that f has a global minimum at (a, b). If this inequality

holds for (x, y) 6= (a, b) sufficiently close to (a, b) then we say that f has a local minimum (or simply a

minimum) at (a, b).

Similarly, if

f(x, y) − f(a, b) < 0,

for all (x, y) 6= (a, b) in the domain of f then we say that f has a global maximum at (a, b) and if this holds

for (x, y) 6= (a, b) sufficiently close to (a, b) then we say that f has a local maximum (or simply a maximum)

at (a, b).

A maximum or a minimum value is called an extremum. The word extrema is the plural of extremum.

Answer 2

Answer:

1/2 is the answer

Step-by-step explanation:

1/2 is the answer