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# Finding the maximum point watch

1. http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN09.PDF

I am stuck on question 5.

I've found dy/dx but unsure how to find the maximum point. I've made dy/dx=0 but don't know what to do from there.

Thankyou!
2. look at the mark scheme
3. right.

y=15x^3/2 - x^5/2
dy/dx=30x^1/2 - 2.5x^3/2
0= 30x^1/2 - 2.5x^3/2
0=12x^1/2 - x^3/2
0=(12 - x)x^1/2
0=12-x
x=12
y= 124.71 (2 d.p.)

I think this is the method. Any comments welcome .
4. (Original post by N!*)
http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN09.PDF

I am stuck on question 5.

I've found dy/dx but unsure how to find the maximum point. I've made dy/dx=0 but don't know what to do from there.

Thankyou!
equate that dy/dx to zero and find x
then to find whether that x gives max or min substitute in second derivative ,if that answer gives a value less than zero then u can confirm it gives a max point
so now to find that maximum point substitute x in ur normal equation and find y
so now the max point is x,y

hope you understand,if not quote me
5. (Original post by hassan_s)
right.

y=15x^3/2 - x^5/2
dy/dx=30x^1/2 - 2.5x^3/2
0= 30x^1/2 - 2.5x^3/2
0=12x^1/2 - x^3/2
0=(12 - x)x^1/2
0=12-x
x=12
y= 124.71 (2 d.p.)

I think this is the method. Any comments welcome .
Don't post full solutions.
6. (Original post by ziedj)
Don't post full solutions.
Sorry, only trying to help...
7. (Original post by ziedj)
Don't post full solutions.
Especially incorrect solutions
8. (Original post by hassan_s)
right.

y=15x^3/2 - x^5/2
dy/dx=30x^1/2 - 2.5x^3/2
0= 30x^1/2 - 2.5x^3/2
0=12x^1/2 - x^3/2
0=(12 - x)x^1/2
0=12-x
x=12
y= 124.71 (2 d.p.)

I think this is the method. Any comments welcome .
the derivation you found is wrong
you should get 9 for your x
but the metod is right
and also you have to find the second derivative to confirm that it gives a maximum point
9. (Original post by gudluck2624)
and also you have to find the second derivative to confirm that gives a maximum point
I think you should get full marks without finding the second derivative. The question specifically says that the diagram shows part of the curve with a maximum point, which implies it's the co-ordinates you obtain will be maximum.
10. (Original post by gudluck2624)
equate that dy/dx to zero and find x
then to find whether that x gives max or min substitute in second derivative ,if that answer gives a value less than zero then u can confirm it gives a max point
so now to find that maximum point substitute x in ur normal equation and find y
so now the max point is x,y

hope you understand,if not quote me
Right, well the bit i'm stuck on is the first part, finding was x is.

I've got

On the mark scheme it's

EDIT: and yeah, what the above poster said, I don't have to find the 2nd derivative.

Don't understand how to get it to that though
11. (Original post by N!*)
Right, well the bit i'm stuck on is the first part, finding was x is.

I've got

On the mark scheme it's

Don't understand how to get it to that though
Try removing a factor of from
12. (Original post by N!*)
Right, well the bit i'm stuck on is the first part, finding was x is.

I've got

On the mark scheme it's

EDIT: and yeah, what the above poster said, I don't have to find the 2nd derivative.

Don't understand how to get it to that though
They have just simplified your equation thats all. Multiply the bracket out and you'll see its the same. You will not need to do that in an exam unless they ask for it or if you are giving a final answer.
13. (Original post by N!*)
Right, well the bit i'm stuck on is the first part, finding was x is.

I've got

On the mark scheme it's

EDIT: and yeah, what the above poster said, I don't have to find the 2nd derivative.

Don't understand how to get it to that though

FROM WHAT you GOT
after simplifying(2 cancels off an so onnn)
9x^1/2 =x^3/2
now bring x^1/2 to the other side so you get x
9= x^3/2 / x^1/2 =>x=9
Hope you understand this
14. (Original post by milliondollarcorpse)
I think you should get full marks without finding the second derivative. The question specifically says that the diagram shows part of the curve with a maximum point, which implies it's the co-ordinates you obtain will be maximum.
yep you dont need to find the second derivative...i didnt actually read the question ...........i just said what are the general points to find a maximum point
15. (Original post by N!*)
Right, well the bit i'm stuck on is the first part, finding was x is.

I've got

On the mark scheme it's

EDIT: and yeah, what the above poster said, I don't have to find the 2nd derivative.

Don't understand how to get it to that though
At a maximum point, Gradient is going to be 0, as it is a stationairy point on the curve

And remember what test to do to find out what a 2nd derivative is (for practice, even if it is implied)

If you can't find it:

Spoiler:
Show

(Differentiate)

dy/dx = 22x^0.5 - 2.5x^1.5

(Set function equal to 0)

22.5x^0.5 - 2.5x^1.5 = 0

(Divide by 2.5)

9x^0.5 - x^1.5 = 0

(Factorise)

(9 - x)x^0.5 = 0

(Solve)

x = 9 or 0

(Test using d2y/dx2 for 9)

4.5x^-0.5 - 1.5x^0.5 = -3

(Conclude)

Therefore, as it is below 0, it is a maximum

(Sub x = 9 back into original equation)

y = (15 * 9^3/2) - (9^2.5) = 162

(Conclude again)

Max point is at (9,162)

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