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

Question

anonymous · Answer

$$A(\Theta)=49\sin 2(\Theta)$$

amistre64 · Answer

you know how to do derivatives?

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

yes, 2pi

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

yes

amistre64 · Answer

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

anonymous · Answer

yes

amistre64 · Answer

like this

anonymous · Answer

yes...I understand that

amistre64 · Answer

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 · Answer

and the biggest sin you can get is 1

anonymous · Answer

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

anonymous · Answer

and that is the equation for the area

amistre64 · Answer

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 · Answer

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

anonymous · Answer

sorry about that

amistre64 · Answer

like this?

amistre64 · Answer

like this?

anonymous · Answer

exactly!

amistre64 · Answer

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

anonymous · Answer

haha...wow...sorry

amistre64 · Answer

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

anonymous · Answer

near the center

amistre64 · Answer

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

amistre64 · Answer

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 · Answer

so would my dimension them be 7 by 14?

amistre64 · Answer

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 · Answer

how did you get these?

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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 · Answer

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 · Answer

thanks!

amistre64 · Answer

its calculus....but it works :)

amistre64 · Answer

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 · Answer

gotcha!!

anonymous · Answer

thanks! have a good night!

amistre64 · Answer

youre welcome :)  Gnite