Question 9.b.iv was bogus! (1 Viewer)

[jiminymacca]

They pretty much give you the equation C=1000 (5-x+2.6 sqrt(x^2+9)) and I assumed you have to find the first derivative of it to get the stationary points and ultimately the minimum, but when I tried to differentiate and solve for dC/dx = 0 I ended up with x^2+9=13/5, which does not have any real solutions! So what happens now? Did I do something wrong??

 

[possessionless]

i got a proper answer for it. it worked for me :S

 

[boxhunter91]

i got a proper answer for it. it worked for me :S

Worked for me aswell.
Cant remember what i got though..

 

[possessionless]

Worked for me aswell.
Cant remember what i got though..

i think i got something like 1.25 but im not entirely sure. something like that.

 

[sinophile2]

Yeah same here, I got that.

 

[kcrocks]

i think i got something like 1.25 but im not entirely sure. something like that.

I got that too

 

[domo.kun]

Yep, got 1.25 as well.

 

[shawty-snappin]

yep 1.25.

and for part v was it directly from P to S?

 

[Thecorey0]

yep 1.25.

and for part v was it directly from P to S?

Yes.

 

[AnandDNA]

woah how did u guys differentiate it

 

[boxhunter91]

woah how did u guys differentiate it

Woot I got 1.25 aswell.

 

[Tumnus]

ooh looks like i got that part right...screwwd another one over though cause i differentiated wrong in the very first step:mad1: corrected it in the last minutes and could finish the part cause tick tock tick tock

d/dx root(3x) = DAMN it i just realised i did that wrong! first one was right....?

DAMN IT, maths i liked you....somewhat

EXAMS

 

[sydneyroosters]

woah how did u guys differentiate it

C = 5000 - 1000x + 2600(x^2 + 9)^1/2

dC/dx = - 1000 + 1300(x^2 + 9)^(-1/2) x 2x
= - 1000 + 2600x / (x^2 + 9)^(1/2)

min/max at dC/dx = 0

-1000 + 2600x / (x^2 + 9)^(1/2) = 0
-1 + 2.6x / (x^2 + 9)^(1/2) = 0
2.6x / (x^2 + 9)^(1/2) = 1 ---square both sides
6.76x^2 / (x^2 + 9) = 1
6.76x^2 = x^2 + 9
5.76x^2 = 9
x^2 = 25/16
x = 1.25 or -1.25 but x > 0
therefore x = 1.25


at x = 1, dC/dx = -ve
at x = 1.25 dC/dx = 0
at x = 1.5 dC/dx = +ve

therefore \_/ , minimum at x = 1.25

sub it in and the cost C = $12200

 

[boxhunter91]

c = 5000 - 1000x + 2600(x^2 + 9)^1/2

dc/dx = - 1000 + 1300(x^2 + 9)^(-1/2) x 2x
= - 1000 + 2600x / (x^2 + 9)^(1/2)

min/max at dc/dx = 0

-1000 + 2600x / (x^2 + 9)^(1/2) = 0
-1 + 2.6x / (x^2 + 9)^(1/2) = 0
2.6x / (x^2 + 9)^(1/2) = 1 ---square both sides
6.76x^2 / (x^2 + 9) = 1
6.76x^2 = x^2 + 9
5.76x^2 = 9
x^2 = 25/16
x = 1.25 or -1.25 but x > 0
therefore x = 1.25


at x = 1, dc/dx = -ve
at x = 1.25 dc/dx = 0
at x = 1.5 dc/dx = +ve

therefore \_/ , minimum at x = 1.25

sub it in and the cost c = $12200

fkn yessssssssssssss! Motherfkrssssssssss

 

[sinophile2]

yep 1.25.

and for part v was it directly from P to S?

No. Had to account for the new cable cost. Just replace 1.1 for 3.6 in the derivative eqn.

 

[Ostentatious]

lolol hi 5

 

[sydneyroosters]

No. Had to account for the new cable cost. Just replace 1.1 for 3.6 in the derivative eqn.

If you replaced 2.6 with 1.1, you will find that C = $7325

From P to S directly, the cost will be only root 34 x 1100 = $6414

From P to R , S to R , cost = 5(1000) + 3(1100) = $8300

Therefore cheapest route is from P to S

 

[Thecorey0]

No. Had to account for the new cable cost. Just replace 1.1 for 3.6 in the derivative eqn.

He is right. Once you do that, you get x=root(300/7). But x is between 0 and 5. Hence it does not exist and the cheapest method is straight there.

 

[raniaaa]

i wrote R to S for the cheapest path =/

 

[MOP777]

part v) I got from p to q then q to s, because I did what he said ^^^ replaced the 2.6 with 1.1 in the original equation then differentiated again (but you just replace it in the derivative lol)