Question

PLEASE HELP?!?! :))
Find the x and y values at the minimum of 2x^2+xy+2y^2+3y+2
And find the x and y values at the maximum of
-3x^2+2xy-4y^2+2y+7

Answer 1

What does min and max mean in this context?

Answer 2

You will need partial derivative for both x and y.

I'll do y.

x+4y+3

You do x.

Answer 3

4x+y

Answer 4

Find partial derivative with respect to x and then y and set it to 0. Solve for x and y.
partial derivative for x is: 4x+y = 0
partial derivative for y is: x+4y+3 = 0

solve for x and y.

Answer 5

for the first equation I got x=0 and y=0, and for the second one I got x=-3 and y=0

Answer 6

you solve the equations simultaneously.

Answer 7

x=0 and y=-3/5

Answer 8

I am getting (1/5, -4/5)

Answer 9

i see, can we go through the second one now

Answer 10

can you verify the answer to the first one?

Answer 11

can you verify the answer to the first one? does the book give you the answer?

Answer 12

i believe it is correct, it is

Answer 13

for the second part I got -6x+2y and 2x-8y+2

Answer 14

yes.

Answer 15

now you set those equations equal to each other?

Answer 16

each expression set to 0.

Answer 17

i got y=3/11 and x=36/11

Answer 18

can you show the work?

Answer 19

nvm i got x its .09091

Answer 20

I am getting x = 1/11 and y = 3/11

Answer 21

Strictly speaking you are supposed to take the second partial derivative and prove these points are minimum/maximum. But that is up to you.

Answer 22

I am getting (1/5, -4/5)

You do NOT have a minimum or maximum established.

You MUST find the three second derivatives and make sense of it.

Answer 23

I am getting x = 1/11 and y = 3/11

You do NOT have a minimum or maximum established.

You MUST find the three second derivatives and make sense of it.

I did all the work on the 1sty Derivatives. It's your turn. Please demonstrate the THREE second derivatives.