Question

Shortest ladder

Answer 1

and?

Answer 2

and?

Answer 3

similar triangles, x/10=4/y

Answer 4

\[z=\sqrt{(4+\frac{ 40 }{ x}) + (x+10)^2}\]

Answer 5

is this correct?

Answer 6

^2 for the first term

Answer 7

i tried to simplify it and differentiate bt it gets complicated ):

Answer 8

Good job, your work looks right so far!

But, instead of differentiating L, why not differentiate L^2? When L^2 is a minimum, L is a minimum... The square root complicates things, and makes zero difference in the end :)
\[\Large L^2 =(4+\frac{ 40 }{ x}) ^2 + (x+10)^2\]chain rule:
\[\Large \frac{ d L^2 }{ dx } = 2\left( \frac{ -40 }{ x^2} \right)(4+\frac{ 40 }{ x}) + 2(x+10)\]

Answer 9

simplifying the above\[\Large 0= \frac{ -320}{ x^2} +\frac{ -3200 }{ x^3} + 2x+20\]it is a bit messy... you can simplify a bit by dividing everything by 2, and multiplying everything by x^3... might help \[\Large 0= -160x-1600 + x^4+10x^3\]ugh..just solve for the zeroes on a calculator.: http://www.wolframalpha.com/input/?i=0%3D+-160x-1600+%2B+x%5E4%2B10x%5E3

so x=5.4288

now plug that in to find L! \[\Large L =\sqrt{(4+\frac{ 40 }{ x}) ^2 + (x+10)^2}\]

Answer 10

http://www.wolframalpha.com/input/?i=L%3D%5Csqrt%7B%284%2B%5Cfrac%7B+40+%7D%7B+x%7D%29+%5E2+%2B+%28x%2B10%29%5E2%7D+when+x%3D5.4288

Answer 11

Just to check: http://www.wolframalpha.com/input/?i=minimum+value+for+x%3E0+of+L%3D%5Csqrt%7B%284%2B%5Cfrac%7B+40+%7D%7B+x%7D%29+%5E2+%2B+%28x%2B10%29%5E2%7D

Answer 12

lol make sure to click approximate form or you get the exact form with square roots of cube roots :P

Answer 13

oh my thanks alot! :D

Answer 14

:D well you had done most of the work so far! Hope the rest made sense!