A tough one. Find the largest possible area for A(theta)=49sin(2theta)

- anonymous

A tough one. Find the largest possible area for A(theta)=49sin(2theta)

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- anonymous

\[A(\Theta)=49\sin 2(\Theta)\]

- amistre64

you know how to do derivatives?

- anonymous

haha some but not with trigonometric functions. I am in precalc

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- amistre64

ok...lets try this route; the largest area of any "sin" is a full circle right?

- anonymous

yes, 2pi

- amistre64

then we need to find a circle that has the same "information" is the given equation.

- amistre64

Do you know the area of a sector? when it involves 2 sides and the sin of an angle?

- anonymous

not yet...I think thats in an upcoming chapter.

- amistre64

ok... and maybe I saw this wrong to begin with...consider this:
the area of a triangle is: (1/2) (base) (height) right?

- anonymous

yes

- amistre64

How do we find the height of a triangle when we know an angle?
does sin(t) = y/r?

- anonymous

yes

- amistre64

like this

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- anonymous

yes...I understand that

- amistre64

good,
then we need to determine the value of "t" that will give us the largest area for the triangle....or thats how I understand this problem

- anonymous

and the biggest sin you can get is 1

- anonymous

would it help to tell you that I have a rectangle inscribed inside a semicircle with a radius of 7cm.

- anonymous

and that is the equation for the area

- amistre64

i keep seeing this 2 different ways...for example:
the area of any given sector of a circle is define as (theta)(r^2)/2

- amistre64

it would..... all the information helps :)

- anonymous

sorry about that

- amistre64

like this?

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- amistre64

like this?

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- anonymous

exactly!

- amistre64

yeah....thatd a been helpful to know lol

- anonymous

haha...wow...sorry

- amistre64

which angle is the theta...near the center of up near a corner?

- anonymous

near the center

- amistre64

if I recall correctly, the angle needs to be 45 degrees.
so 2t = 45 t = 22.5

- amistre64

the largest area you can get is formed by a square.... and you need 2 squares side by side to make this the max area under the half circle

- anonymous

so would my dimension them be 7 by 14?

- amistre64

2 sides equal 7sin(45); and the other 2 sides are 28sin(45)
4.95 is one side...
9.90 is the other side......
it gives you an area about equal to 49.005

- anonymous

how did you get these?

- amistre64

magic :) first I killed a live chicken; then a spread its blood around the bedposts......

- anonymous

haha seriosuly!...thats how I feel about these problems! You need to do voodoo to get an answer!!

- amistre64

we take the equation for a half circle:
y = sqrt(7^2-x^2) and the Area of the "rectangle" to be maximized.
A = xy ; substitute y = sqrt(49 - x^2) into this equation

- amistre64

A = x(sqrt(49-x^2))
now find the derivative....
A' = x(-x/(sqrt(49-x^2))) +1(sqrt(49-x^2)) and make that equal to zero.... then solve for x :)

- anonymous

thanks!

- amistre64

its calculus....but it works :)

- amistre64

the other way is just to notice that the largest area that can be produced in a quarter of a circle is a square....a square has a 45 degree angle where we need it down there

- anonymous

gotcha!!

- anonymous

thanks! have a good night!

- amistre64

youre welcome :) Gnite

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