Question

Find the rectangle of largest area that can be inscribed in a semicircle of radius R, assuming that one side of the rectangle lies on the diameter of the semicircle.

NOTE: Let L denote the length of the side that lies on the diameter and H denote the height of the rectangle. Your answer will likely involve R.


L=

H=

Answer 1

it is a calculus problem, in the chapter of optimization

Answer 2

ok

Answer 3

area of that rectangle is \(A=2xy\)

Answer 4

since \(x^2+y^2=r^2\) you have
\[y=\sqrt{r^2-x^2}\]

Answer 5

that makes
\[A(x)=2x\sqrt{r^2-x^2}\] etc

Answer 6

@satellite73 so what would be next?

Answer 7

calculus

Answer 8

take the derivative, find the critical points etc

Answer 9

ok im taking the derivative right nw

Answer 10

(2 (r^2 - 2 x^2))/Sqrt[r^2 - x^2]

Answer 11

you need the product rule for it
maybe... let me check

Answer 12

yeah that looks good

Answer 13

when i make it equal to zero to then find the critical points, i get

r/sqrt(2) and -r/sqrt(2)

Answer 14

set
\[r^2-2x^2=0\] solve for \(x\)

Answer 15

ok

Answer 16

yeah that looks good also

Answer 17

so whats next?

Answer 18

after the critical points

Answer 19

nothing

Answer 20

well i guess i don't mean nothing
lets go back to the picture
you have found \(x=\frac{r}{\sqrt2}\)

Answer 21

you are asked for \(H\) which is evidently
\[\frac{2r}{\sqrt2}=\sqrt2r\]

Answer 22

and also \(L\) which we called \(y\) and you can solve for since you know
\[x^2+y^2=r^2\]

Answer 23

so H is? and L?

Answer 24

i guess i confused you
i said what \(H\) is above

Answer 25

H = sqrt(2)r ?

Answer 26

I answered that for H and i got it wrong

Answer 27

oh because i am an idiot and don't know how to read
L is the length, H is the height (doh)

Answer 28

that makes \(L=\sqrt2 r\)

Answer 29

and H?

Answer 30

@satellite73

Answer 31

I need help finding H, @satellite73

Answer 32

@dan815 help me

Answer 33

@ganeshie8