Really need some help with this one guys. Been stuck several days now. :/

Really need some help with this one guys. Been stuck several days now. :/

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desmos.com/calculator/t8hiwwoin7
imgur.com/a/adATx
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Let x=2*cos(y)
then perimeter(y)=2*sin(y)+4*cos(y)

differentiate function perimeter and set as zero to find max

P(x)=2*(x+sqrt(1-x^2/4))
solve for P'(x)=0 and x=4/sqrt(5)

also this is literally highschool stuff

>sqrt(1-x^2/4)

i-is this bait

Put the semicirle in the cartesian plane

Then the equation for the perimeter is 2x + 2(1 - (x/2)^2)^0.5

Do some calculus

explain user, please :=)

desmos.com/calculator/t8hiwwoin7

I did the problem in demos. Hopefully this will help you find your answer.

n-nevermind

thanks!

The perimeter is given by
[math] f(x,h) := 2x + 2h [/math].
Then you also have the constraint
[math] g(x,h) := \left( \frac{x}{2} \right)^2 + h^2 - 1 = 0 [/math].
So you just have to solve the system:
[math] \nabla f (x,h) = \lambda \nabla g(x,h) \\ g(x,h) = 0[/math].

The solutions are
[math]x = \frac{4}{\sqrt{5}} \\ h = \frac{1}{\sqrt{5}} \\ \lambda = \sqrt{5} [/math].
Which makes the perimeter equal to [math] 2 \sqrt{5} [/math]

also thank me for you underage asshole

lol, sorry m8. Thank you as well you fantastic human being

thanks family

x*sqrt(1-(x/2)^2)

...

prove me wrong

That would be the formula for area, not perimeter.

oops I was driving when I read the problem. ok then how about
2x + 2sqrt(1-(x/2)^2)

seems to look better :)

Do you know nothing about trigonometry?

Let me put it like this, I have a decent grade in trig., don't deserve it.

Literally the best way. This is what I thought,

>grade in trig

wait, is trigonometry a separate class to mathematics?

In the USA it's a college level class taken in the second semester after taking College Algebra in the first semester.

surely some basic stuff is introduced in middle/high school?

at least SOH CAH TOA

Lies

2x+2sin(arccos(x/2))

arccos = cos^-1

from the centre of x to either lower corner the distance is x/2 of course

then x/2 is the adjacent of a right triangle with the radius as the hypotenuse

then use inverse cosine to get the angle of this triangle

use sine with the angle to get the opposite side

now you have the 2 dimensions of the rectangle, so use 2 of each (hence 2x+2...) and you got it

shouldn't be hard to solve for zero

"college level" probably means at a level for people in high school aiming for college instead of university, or at least that's what my area's school system does

UPDATE, please go to:
imgur.com/a/adATx

kek

how would you find minima and maxima in any function of x?

f'(x)=0 ?

yes

dont forget what x represents after your calculations

so

P ’(0) = 2 , 2√1-0^2/4+2*0
Max value of x = 2

Then to find max perimeter i inserted x=2 in the original function

P (2) = 4 , 2*2 + 2(1 - (2/2)2)^(1/2)

The max perimeter is = 4 ?

Is that correct, it feels wrong :/

bump
anyone?

very smooth

one thing to consider is what x can possibly be

how big (and small) could x be for the question to still make sense?

then P is your function of x

so you want to find where P'(x) = 0 and go from there

so

f'(0)=4√5/5

Highest perimeter = 4√5/5 = 1.78

?