Respuesta :
Answer:
the minimum value of f(x,y) is fmin=f(0, ±11)=-121 and the maximum is fmax=f(±11, 0)= 121
Step-by-step explanation:
Since
f(x,y) = x²-y²
subject to the constraint
x²+y²=121 → x² = 121 - y²
f(x,y) = 121 - y² -y² = 121 - 2*y²
f(x,y) =121 - 2*y²
then since f(x,y) diminishes with increasing y , then the minimum f is when y is maximum
since
x²+y²=121 → y² = 121 - x² → y max = √121 , (x =0)
f min = f(0, ±11) x²-y² = 0²-(±11)² = -121
f min = (-121)
for the maximum value
f(x,y) =121 - 2*y²
also since f(x,y) diminishes with increasing y , then the maximum of f can be found when y² is minimum → since positive and negative values increase y² the minimum is y=0. This can be proven from
f(x,y) =121 - 2*y²
df/dy = 2*2*y = 4*y =0 → y=0
also
x²+y²=121 → x² = 121- y² = 121-0² = 121 → x=±11
then the maximum value of f is
f max= f(±11, 0) = 121 - 2*0² = 121
