Question

find the local max/min/saddle points

f(x,y) = (1+xy)(x+y)

Answer 1

where are you stuck?

Answer 2

i can't find the critical points

Answer 3

i get fx = 1 + 2xy + y^2 and fy = 1 + 2xy + x^2 but how do i solve these?

Answer 4

you want fx=0 and fy=0

thus fx=fy

1 + 2xy + y^2=1 + 2xy + x^2

y^2=x^2
\(y=\pm x\)

so if y=x

1 + 2xx + x^2=0
1+3x^2=0 which is not possible

if y=-x

\[1 - 2xx + (-x)^2=0\]

\[1-2x^2+x^2=0\]

\[1-x^2=0\]
\[x=\pm 1\]

so the solution is (x=1 and y=-1) and (x=-1 and y=1)

Answer 5

(x=1 and y=-1) or (x=-1 and y=1)

Answer 6

why can you set fx = fy?

Answer 7

because they are both equal to zero

Answer 8

fx=0
fy=0

thus fx=0=fy
fx=fy

Answer 9

okay hang on, let me see if i understand this

Answer 10

why are the critical points (1,-1) and (-1,1) but not (-1,-1) or (1,1)

Answer 11

if it were (1,1) or (-1,-1) then y=x...and from above we see that that is not possible

Answer 12

ahhh okay!

Answer 13

also try plugging (1,1) into 1 + 2xy + y^2

you get

\[1+2+1=3\ne 0\]

Answer 14

lol


1+2+1=4 which is not 0

Answer 15

okay can you help me on a similar one?

Answer 16

i can ask it in a new question