- #1
dagg3r
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HI GUYS
i just have some hard problems on partial differentiation... hope you guys can help out
1. find "day z / day y" its the funny backwards 6 symbols instead of the normal differntiation symbole
if z=4y^3 * sqrt(x) - ln(4xy^3) + 2x
i did this:
i diffenriated 4y^3 * sqrt(x) with respect to y, and since sqrt(x) stays constant, it remains hence i get
12y^2*sqrt(x) . i then diffed ln(4xy^3) + 2x
i did
ln(u)
u=4xy^3 u`= 12xy^2
=1/u * u`
=(1/4xy^3) * 12xy^2
= 3/y
and the 2x disappears
hence my ultimate answer is
"day z/ day y"=12y^2*sqrt(x) - 3/y
2. FInd the stationary point of
z= 5x^2 + 23x + y^2 + 8y + 3xy + 32 amd determine the nature of this point
day z / day x = 10x + 23 + 3y
day z / day y = 2y + 8 + 3x
use elimination hence
10x + 23 + 3y *2
2y + 8 + 3x * 3
-----------------
20x + 46 + 6y 3)
6y + 24 + 9x 4)
---------------
11x -22 = 0
x = 2 sub this into eqn 3)
y= -43/3
now sub x=2, y=-43/3 into z=
hence z=5(2)^2 + 23(2) + (-43/3)^2 + 8(-43/3) + 3*(2)(-43/3) + 32
z=925/9
hence stationary points i think are (2, -43/3 , 925/9)
ok assuming that's right i have to determine the nature of the point, to my knowledge i would have to
use the formula g= (day^2 z/day*x^2)*(day^2 z/ day*y^2) - ( day^2*z / day*x*day*y )
day^2 z / day^2 x^2 = 10
day^2 z / day^2 y^2 = 2
day^2*z / day*x*day*y = 3
hence g = (10)*(2) - 3
G= 17
HENCE 17 > 0 , AS (day^2 z / day^2 x^2) IS > 0, AND G> 0, THEN THE STATIONARY POINT IS A MINUMUM POINT
I THINK I DID IT WRONG HAVE NO IDEA
i just have some hard problems on partial differentiation... hope you guys can help out
1. find "day z / day y" its the funny backwards 6 symbols instead of the normal differntiation symbole
if z=4y^3 * sqrt(x) - ln(4xy^3) + 2x
i did this:
i diffenriated 4y^3 * sqrt(x) with respect to y, and since sqrt(x) stays constant, it remains hence i get
12y^2*sqrt(x) . i then diffed ln(4xy^3) + 2x
i did
ln(u)
u=4xy^3 u`= 12xy^2
=1/u * u`
=(1/4xy^3) * 12xy^2
= 3/y
and the 2x disappears
hence my ultimate answer is
"day z/ day y"=12y^2*sqrt(x) - 3/y
2. FInd the stationary point of
z= 5x^2 + 23x + y^2 + 8y + 3xy + 32 amd determine the nature of this point
day z / day x = 10x + 23 + 3y
day z / day y = 2y + 8 + 3x
use elimination hence
10x + 23 + 3y *2
2y + 8 + 3x * 3
-----------------
20x + 46 + 6y 3)
6y + 24 + 9x 4)
---------------
11x -22 = 0
x = 2 sub this into eqn 3)
y= -43/3
now sub x=2, y=-43/3 into z=
hence z=5(2)^2 + 23(2) + (-43/3)^2 + 8(-43/3) + 3*(2)(-43/3) + 32
z=925/9
hence stationary points i think are (2, -43/3 , 925/9)
ok assuming that's right i have to determine the nature of the point, to my knowledge i would have to
use the formula g= (day^2 z/day*x^2)*(day^2 z/ day*y^2) - ( day^2*z / day*x*day*y )
day^2 z / day^2 x^2 = 10
day^2 z / day^2 y^2 = 2
day^2*z / day*x*day*y = 3
hence g = (10)*(2) - 3
G= 17
HENCE 17 > 0 , AS (day^2 z / day^2 x^2) IS > 0, AND G> 0, THEN THE STATIONARY POINT IS A MINUMUM POINT
I THINK I DID IT WRONG HAVE NO IDEA