Question

maximise problem

Answer 1

i cant do 12d

Answer 2

how can you use 11 metres when theres only 10 metres of string?

Answer 3

So for part d, one piece measures between 0 and 11, and the total string is still 10?
Yah that's strange...

Answer 4

yes i have no idea i'll upload solution doesnt make sens ehtough

Answer 5

hi??

Answer 6

Still trying to figure it out c: gimme few mins.

Answer 7

@jim_thompson5910 @Hero @phi
I'm paging a few of the smarty pants fellas on here.
Maybe one of them can come take a look and figure out what's going on.

I'm stumped, sorry to say. :(

Answer 8

okay cheers. the answer might be wrong? not sure

Answer 9

@mathslover

Answer 10

What do you mean you can't do it. It looks like it's already been done....

Answer 11

Explain to me exactly what you're trying to do with 12d.

Answer 12

how is x beytween 0 11 isnt there only 10 metres of string

Answer 13

I don't see where x = 10......

Answer 14

Bro, where do you see that 0 < x < 11 ...because I don't see that anywhere. Especially for part 12d. What I see is that A = 3.625^2 and x = 7. Please explain to me your confusion regarding this.

Answer 15

If two squares are formed but xE[0,11]
??

Answer 16

how to get a=3.626^2 and x=7

Answer 17

@Hero you musta clicked the wrong link. Are you only looking at the solutions link?

Answer 18

Yes, I am. I suppose the solutions link is the wrong link?

Answer 19

Well we're trying to undestand how they came to that answer :d
The question doesn't seem to make sense in part d.

Maybe we're misinterpreting x some how.

Answer 20

Looks like it should be \[x \in [0,10] \]. Probably just a typo, but nothing to get overly panicky about.

Answer 21

dA/dx = x-5 / 4 its not increasing for [4,7] ??

Answer 22

You're right. It's increasing [5,10]. Okay, so now that you know that, fix the mistakes.

Answer 23

so Amax=A(10) why do both answers say 29/8

Answer 24

Bro, you have the correct answer now. I don't know why they printed the incorrect answer, but don't stress over it. Go over it with your teacher tomorrow or Monday.

Answer 25

okay fanks

Answer 26

From the answer, it seems they meant to ask in 12d for
x in the interval [4,7] (not [0,11] )

the parabola is symmetric about x=5


this parabola has a min at x=5, so its max will be one side or the other of the interval given. If you notice that 4 is close to 5 (the min point) and 7 is further away, you can say 7 will be the max point for the interval [4,7]

If it was not clear, you would try 4 and 7 in the equation in part (a) to find the one with the biggest area.