Question

Every second counts. Optimization

Answer 1

do i use the formula d=vt to find the function for T?

Answer 2

Yes. The swim distance is (from pythagoras) \[\large \sqrt{50^2+x^2}\] the time it takes you to swim that distance is then that divided by his speed 2.

The running distance is 50-x, the time is then (50-x)/4 since he can run at 4m/s

T is the sum of those.

Answer 3

okay for part b, its u differentiate that equation =0 ?

Answer 4

hm so for b, x= sqrt 20?

Answer 5

you have T as\[\Large T=\frac{ 1}{2 }\sqrt{50^2+x^2}+\frac{ 1 }{ 4 }(50-x)\]right?

Answer 6

yup!, i differentiated that and i got \[dT/dx =\frac{ 1 }{ 2 }x(50^2+x^2)^\frac{ -1 }{ 2 }-\frac{ 1 }{ 4 }\]

Answer 7

then i equate that to 0, then get x=sqrt 20

Answer 8

derivative of x^2 is 2x, not x

Answer 9

i used chain rule, so have to bring down the power too right? itll cancel out

Answer 10

Hmm, yeah you're right. But your x isn't

Answer 11

oh okay i think i found my mistake its suppose to be x=sqrt 100/3?

Answer 12

you might have missed a 1/2... http://www.wolframalpha.com/input/?i=minimum+y%3D%5Cfrac%7B++1%7D%7B2+%7D%5Csqrt%7B50%5E2%2Bx%5E2%7D%2B%5Cfrac%7B+1+%7D%7B+4+%7D%2850-x%29

Answer 13

after setting it =0 and rearranging a bit:\[\Large 0.25(50^2+x^2)^\frac{ 1 }{ 2 } =\frac{ 1 }{ 2 }x\]then square both sides

Answer 14

how did you get that T^T

Answer 15

\[\Large 0 =\frac{ 1 }{ 2 }x(50^2+x^2)^\frac{ -1 }{ 2 }-\frac{ 1 }{ 4 }\]add 1/4 to both sides, multiply both sides by (50^2+x^2)^(1/2)

Answer 16

okay so resulting would be \[4x^2=50^2+x^2\] am i right?

Answer 17

omygosh, i think i know why im so confused lmao, 50^2 is not 100 i kept thinking it was, anyways yeah the answer shd be 50/sqrt3 haha

Answer 18

then how do i go with part c? differentiate it again?

Answer 19

Yep. Haha i wondered if i should remind you that 5-^2 is 2500... :P

Answer 20

c is plugging values into T.

Answer 21

x=0, x=50/sqrt3, x=50, check that the min is indeed a min (it's a global minimum so it will be)

Answer 22

its a positive so it means its a minimum right?

Answer 23

No, just find the value of T at each point.

Answer 24

The minimum will be the... minimum value :P

Answer 25

oh.. okay so it is minimum at x=0? i got 12.5, ~23 and ~35

Answer 26

\[\Large T=\frac{ 1}{2 }\sqrt{50^2+x^2}+\frac{ 1 }{ 4 }(50-x) \]you're using this, right?

Answer 27

yes

Answer 28

Check your work for x=0 then... should be 25+12.5 so 37.5

Answer 29

no i might have messed up my values again haha

Answer 30

yeah.. omygosh..

Answer 31

You can prob do x=0 in your head if you try :)

Answer 32

i minused it sigh

Answer 33

It happens!

Answer 34

okay so min at x=50/sqrt3?

Answer 35

graphing this need a software right? or can graph it manually? haha

Answer 36

Yep. Looking at a graph can help in these kinda questions, while finding the points, to make sure your values make sense.
http://www.wolframalpha.com/input/?i=T%3D%5Cfrac%7B++1%7D%7B2+%7D%5Csqrt%7B50%5E2%2Bx%5E2%7D%2B%5Cfrac%7B+1+%7D%7B+4+%7D%2850-x%29

use a graphing calculator. Waste of time by hand.

Answer 37

i see, okay thanks! (:

Answer 38

http://www.wolframalpha.com/input/?i=T%3D%5Cfrac%7B++1%7D%7B2+%7D%5Csqrt%7B50%5E2%2Bx%5E2%7D%2B%5Cfrac%7B+1+%7D%7B+4+%7D%2850-x%29+for+%5B0%2C+50%5D

better graph