Find the absolute maximum and minimum of the function f(x, y)=x2−y2f(x, y)=x2−y2 subject to the constraint x2+y2=121x2+y2=121. as usual, ignore unneeded answer blanks, and list points in lexicographic order. absolute minimum value

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

step-by-step explanation:

no, her answer is not correct.   we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306.   this is not equal to the square of the longest side ( that square being 395.   answer c is correct.