Fun Trig question for math freaks(VERY HARD) (1 Viewer)

[chrisk]

Minimum value for the desk

download the diagram from the attachment.

some are actually confused so you must download the picture.

Two corridors meet at right angles and are 2m and 3m wide respectively.
Θ is the angle marked on the given figure and ABCD is a desk
which must be kept horizontal as it moves around the corner from one
corridor to the other without bending it.

FIND the greatest possible area of the desk so that it can be horizontally carried around the corners.

make sure you use calculus to find the minimum value. This involves inverse function as well.

I don't know the answer for it so please discuss.

thanks!

 

[youngminii]

download the diagram from the attachment.

Two corridors meet at right angles and are 2m and 3m wide respectively.
Θ is the angle marked on the given figure and ABCD is a desk
which must be kept horizontal as it moves around the corner from one
corridor to the other without bending it.

FIND the greatest possible area of the desk so that it can be horizontally carried around the corners.

Sooo basically the question is what is the biggest possible area of the desk so that it can be placed in the 2m place.


I don't know the answer for it so please discuss.

thanks!

Actually, that's not what the question's asking. Think about it. The desk cannot be bent in any way, so it has to be kept rigid as it turns the corner.

I believe the answer is 6m^2. If I'm right, this is far from Extension 2 Mathematics and it most certainly does not use any trig at all.

 

[chrisk]

can you show me the working how to got 6m^2?
i think you misunderstood something

you need to use trig to find the length and the width of the desk and then
this involves trig because you should use the Θ.
Then you need to find dA/dΘ to find the minimum possible value of the area.
you let dA/dΘ=0

 

[youngminii]

can you show me the working how to got 6m^2?
i think you misunderstood something

you need to use trig to find the length and the width of the desk and then
this involves trig because you should use the Θ.
Then you need to find dA/dΘ to find the minimum possible value of the area.
you let dA/dΘ=0

Oh I see.. In that case, hmm..