Help with 2011 Exam 2

[m.arcus]

Exam here: http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011mmcas2-w.pdf
Solutions here: http://www.vcaa.vic.edu.au/Documents/exams/mathematics/mmcas2_assessrep_11.pdf

Could someone explain question 4f to me please? I don't understand why the solution is obtained through this way.

P/s: this exam was freaking hard, I'm worried for this year's now tbh

[abeybaby]

this could help


https://www.box.com/files#/files/0/f/147488137/1/f_1111258089

[m.arcus]

Sorry, but I am unable to view the file even though I signed up and all

Could you please upload it somewhere else?

[abeybaby]

It's too large to attach here, is there an email address I can send it to?

[BubbleWrapMan]

There is a certain value of k for which the minimum time is achieved by running directly to the desalination plant, which you can find by letting x = √(7)/2 and solving dT/dx = 0 for k, and you get k = 5√(37)/74. If k is any bigger he's still better off running straight to the plant, so k can be anything greater than or equal to that value.