The Student Room Group

Differentiation c2

we've jsut gotta teach us this ourselves, so im having some difficulties. i found differentiation in c1 relatively easy, but getting confused on this, because im just not sure of the methods, if im shown how to do this question maybe i'll be able to go on from there

Given that y=x^3/2 + 48/x (x > 0)

a) Find x and why when dy/dx = 0

b) Show that the value of y which you dound in A is a minimum.


i differentiate the equation, but im not sure what to do next :s-smilie:

thanks :smile:
Reply 1
Set dy/dx=0 and solve for x, then plug in the value you found in the equation to find y. Now find d^2x/dy^2 and plug in the value of y you found. Remember that if d^2y/dx^2 > 0 then we have a minimum.
Reply 2
Hello!

y=0.5x^3 + 48x^-1
dy/dx = 1.5x^2 - 48x^-2 (sorry, 1.5 is just easier/less ambiguous to type than 3/2 x...)

Then put dy/dx = 0 for a stationary point (either a minimum or a maximum)
so
1.5x^2 - 48/x^2 = 0
1.5x^2 = 48/x^2
3x^4 = 96
x^4 = 32 so x=2 or x=-2

If x=2, then y=2^3 / 2 + 48/2 = 28 (x=2, y=28)
If x=-2 then y=-2^3 /2 + 48/-2 = -28 (x=-2, y=-28)


If the point is a minimum, then (as dvs said) the second derivative (differentiate dy/dx) is positive (greater than 0)
so
dy/dx = 1.5x^2 - 48x^-2
Differentiate that
d2y/dx^2 = 3x + 96/x^3

When x=2, d2y/dx^2 = 3(2) + 96/(2^3) = 6 + 12 = 18 >0
So when x=2, the stationary point (2, 28) is a minimum.

The other stationary point, when x=-2,
d2y/dx^2 = 3(-2) + 96(-2^3) = -6 - 12 = -18<0
So when x=-2, the stationary point (-2, -28) is a maximum.

Um.
HANG on a minute!
that just SO doesn't make sense, I am a complete idiot but have just re-done the question three times and I still get the same answers.

Aha! I've drawn the graph and it has two branches (well, duh, one day I'll get a brain) and the left-hand one is an n shape, so the x=-2 is the maximum) and the right-hand one is a U shape, so the when x=2, it's a minimum (which I suppose just seemed weird given the co-ordinates)

Um, yes. Sorry about all the wibbling!
the solution is actually right, the question's weird. Honest. :blush: Sorry!

love danniella
Reply 3
danniella
x^4 = 32 so x=2 or x=-2

You might want to recheck that.
Reply 4
Hi, Danniella hey Grant.

Okay

a)

y = x^3/2 + 48/x

dy/dx = 3/2x^1/2 - 48x^-2

if dy/dx = 0

3/2x^1/2 - 48x^-2 = 0
3/2x^1/2 = 48x^-2
x^1/2 = 32x^-2
x^1/2/x^-2 = 32
x^5/2 = 32
x^1/2 = 2
x = 4

b)

Sub x = 4 into y = x^3/2 + 48/x

y = 20

x = 4 y = 20

d^2y/dx^2 = 3/4x^1-2 + 96x^-3

sub x = 4 into 3/4x^1-2 + 96x^-3

d^2y/dx^2 = 15/8 therefore minimum.

Grant some rep would be nice :P
Reply 5
Given that y=x^3/2 + 48/x (x > 0)

dy/dx= 1.5x^0.5 - 48/x^2
at dy/dx=0 =>1.5x^0.5 - 48/x^2=0
1.5x^2.5=48
x^2.5=32
x=4

b) Show that the value of y which you dound in A is a minimum.

d²y/dx²=0.75x^-0.5 + 96/x^3
at x=4
d²y/dx²=15/8> 0 therefore minimum

Just did it as well since there were two different answers.
Reply 6
Lol, i always do it the long way! I checked in the back of the book so i know my answers were right.
Reply 7
why r there that many answers? it's just a question! stop posting new answers!
Reply 8
dvs

x^4=32
x=+2 or x=-2
You might want to recheck that.


:bangs head on desk:
Someone shoot me now. :redface:
I think I might want to re-do the question yet again.
Preferably by reading it correctly as x^(3/2) as well...
I suppose I should just be REALLY grateful I didn't present those answers in class!

love danniella
Reply 9
yazan_l
why r there that many answers? it's just a question! stop posting new answers!


Mine and Malik's were exactly the same and the right answer so its not a new answer it is the answer.

Shoots Danniella for answering half the question on this board :P Danniella you seem to know your stuff lol dont beat yourself up.