Partial differention hard questions

dagg3r

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 thats 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

mathmike

isnt der of z wrt y of sqrt(x) = 0

HallsofIvy

Yes, it is that's why the derivative of [tex]4y^3\sqrt{x}[/tex] is
[tex]12y^2\sqrt{x}[/tex] just as dagg3r says- even if you use the "product rule". What did you think the derivative should be?
The partial derivative of [tex]z= 4y^3\sqrt{x} - ln(4xy^3) + 2x[/tex]
with respect to y is:
[tex]\frac{\partial z}{\partial y}= 12y^2\sqrt{x}- \frac{1}{4xy^3}\left(12xy^2)\right)[/tex]
= [tex]12y^2\sqrt{x}- 3y[/tex] exactly what dagg3r had!
(Click on the TEX above to see the code I used- your "day" is driving me mad! I would have preferred you just use "d" with the notation that it is partial differentiation.)
Okay, you got
[tex]\frac{\partial z}{\partial x}= 10x+23+ 3y[/tex]
and
[tex]\frac{\partial z}{\partial y}= 2y+ 8+ 3x[/tex]
Of course, at a stationary point, the partial derivatives are 0:
10x+ 3y+23= 0 and
3x+ 2y+ 8= 0 (You did not write "= 0". While I understood what you were doing, the "= 0" makes it clearer. Also I have written the two equations with x first, then y. That also makes it a little simpler to see what to do.
Yes, multiply the first equation by 2 and the second by 3, to get "6y" in both and then subtract:
20x+ 6y+ 46= 0
9x+ 6y+ 24= 0
which gives
11x+ 22= 0.
You subtracted wrong! (That might be because of the way you had written the equations.)
x= -2, not 2, and then, from the second initial equation, -6+ 2y+ 8= 0 so
2y= -8+ 6= -2, y= -1. The stationary point is at (2, -1).
z(2, -1)= 5(4) + 23(2) + (-1)^2 + 8(-1) + 3(2)(-1) + 32
= 20+ 46+ 1- 8- 6+ 32= 85.
NOW determine the nature of the stationary point.