# rate of change

• May 1st 2007, 08:14 AM
partial diff. min, max & saddle
Hey All,

how do determine the position and nature (i.e minimum, maximum or saddle point) of the stationary points of the function.

f(x,y)=x^2 - 4xy + y^3 + 4y

this is what I have done so far
dz/dx= 2x-4y
dz/dy=-4x+3y^2+4

d^2z/dxdy=-4

how do you complete the question?

thanks
• May 1st 2007, 09:35 AM
TD!
The stationary points are the solutions to grad(f) = 0.
Use your partial deratives, this gives a system in x and y.
Use this test to determine the nature of these points.
• May 1st 2007, 10:16 AM
sorry wrong title! thanks for the link. cool site
• May 1st 2007, 10:20 AM
TD!
I thought the title was kind of strange ;)

That site is a very useful reference for math topics.
• May 2nd 2007, 10:21 AM
Hi

so do i make the partial diff. equal to zero then find x?

is it the same rule for normal differentation too?

thanks
• May 2nd 2007, 12:24 PM
Jhevon
Quote:

Originally Posted by dadon
Hi

so do i make the partial diff. equal to zero then find x?

is it the same rule for normal differentation too?

thanks

yes, but you have to find coresponding y's as well, you need coordinates for each critical point to use in the formula
• May 2nd 2007, 09:03 PM
thanks

what are the general rule for partial diff. to get min max and saddle

(also in saddle the same as inflexion?)

so you make the partial diff to zero.. to get the stationary points.

then you differentiation again ... now this is where i get lost

how do u find the max, min saddle from second partial diff. ?

what is the general rule?

thanks
• May 2nd 2007, 11:17 PM
Jhevon
Quote:

Originally Posted by dadon
thanks

what are the general rule for partial diff. to get min max and saddle

(also in saddle the same as inflexion?)

so you make the partial diff to zero.. to get the stationary points.

then you differentiation again ... now this is where i get lost

how do u find the max, min saddle from second partial diff. ?

what is the general rule?

thanks

the website Ticbol posted tells you everything you need to know:

find fx and fy and set them equal to zero to find the critical points.
find also fxx and fyy and fxy.

FOR EACH CRITICAL POINT, evaluate fxx, fyy, fxy and plug them into the formula D = fxx*fyy - (fxy)^2, and the classifications are as follows:

if D > 0 and fxx> 0 at the specific point, it is a relative min
if D > 0 and fxx< 0 at the specific point, it is a relative max
if D < 0 at the specific point, it is a saddle point
if D = 0, you need another test, this one is inconclusive
• May 3rd 2007, 05:53 AM
Soroban

Have you never done one of these before?

Quote:

Determine the position and nature (i.e minimum, maximum or saddle point)
of the stationary points of the function: .f(x,y) .= .x² - 4xy + y³ + 4y

Find the two first partial derivatives, equate to zero, and solve the system.

. . f
x .= .2x - 4y .= .0 - - - - . [1]
. . f
y .= .-4x + 3y² + 4 .= .0 .[2]

From [1], we have: .x = 2y . [3]

Substitute into [2]: .-4(2y) + 3y² + 4 .= .0

. . We have a quadratic: .3y² - 8y + 4 .= .0

. . which factors: .(3y - 2)(y - 2) .= .0

. . and has roots: .y .= .2/3, 2

Substitute into [3]: .x .= .4/3, 4

The stationary points are: .(4/3, 2/3, 32/27) and (4, 2, 0)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Second Partials Test

We need: .D .= .(f
xx)(fyy) - (fxy

We have: .f
xx = 2, . fyy = 6y, . fxy = -4

Hence: .D .= .(2)(6y) - (-4)² .= .12y - 16 .= .4(3y - 4)

At (4/3, 2/3, 32/27): .D .= .4[3(2/3) - 4] .= .-8 . . . negative: saddle point

At (4, 2, 0): .D .= .4(3·2 - 4) .= .+8 . . . positive: max or min
. . f
xx = 2 . . . positive: concave up . . . minimum

• May 3rd 2007, 09:39 AM