Question

Calculus Problem-Optimization Problem
A rectangle is inscribe in a circle of radius 2m. Find the maximum area of this rectangle.

Answer 1

I need to draw it

Answer 2

Yes. I wish the drawing tool worked.....

Answer 3

ok well basically it is a rectangle made up of 2 rectangles. The dimensions for each rectangle is x and y

Answer 4

I have to go eat lunch, but the answer should come out to be a square. That's what generally has the largest surface area in these optimization problems.

Answer 5

but I need to give the set up. I believe the solution is 10 x 10 if your advice works out since that is the only logical solution on my puzzle worksheet

Answer 6

Can you post the diagram here?

Answer 7

Not sure, I would have to scan it in.

Answer 8

Remember that the diagonal of the rectangle must be less than or equal to the diameter of the circle.

Answer 9

not sure this is what your on about, but its a square inside a circle? draw a cross with 90 degree angles between each line, and then join the ends of each line (where the radius' meet the circumference and you get the biggest square possible

Answer 10

http://www.twiddla.com/1471081 have fun

Answer 11

@Agent48 If I draw it there, can I post it here?

Answer 12

http://www.twiddla.com/1471081

Answer 13

I would say
L^2 + W^2 = 4^2
A = L * W
A = L * sqrt(4^2 - L^2)

Then you maximize A based on L.

Answer 14

Ok that is the best I can do to draw it.
My primary equation is A=x^2y
I am having problems getting my secondary equation

Answer 15

Why would it be x^2y?

Answer 16

because the maximum area is the rectangle and that is the primary equation

Answer 17

A=2x(y) I forgot that is A=lw

Answer 18

Okay, so you are making x be half the width?

Answer 19

I am given a rectangle that is made up of 2 rectangles. The dimension of each rectangle is x and y. I wish the drawing tool was working

Answer 20

Length is x + x and width is y

Answer 21

Ok

Answer 22

I am stuck on the secondary equation

Answer 23

I thought the secondary should have to do with the circle since the rectangle is inscribed in the circle of radius 2

Answer 24

It would be the fact that the diagonal of the whole rectangle less than or equal to the diameter or the circle.

Answer 25

Seems the equation tool is not working as well

Answer 26

Latex wouldn't show up anyway.

Answer 27

A=4x time the square root of (4-x^2)

Answer 28

How did you get that?

Answer 29

From the puzzle worksheet, it is the only solution that works (ie letter works)

Answer 30

What were your first two equations?
A = 2x y and?

Answer 31

Primary is A=(2x)y but I need a secondary to get rid of x or y

Answer 32

I need to go scan these pictures and repost. Let me go do that. I will repost this.