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.
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
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
Hello, dadon!
Have you never done one of these before?
Find the two first partial derivatives, equate to zero, and solve the system.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
. . fx .= .2x - 4y .= .0 - - - - . [1]
. . fy .= .-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 .= .(fxx)(fyy) - (fxy)²
We have: .fxx = 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
. . fxx = 2 . . . positive: concave up . . . minimum