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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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
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
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
?