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A Problem for Fermat

Find (x, y) that maximises

f(x, y) = | (cos x)^4 + (cos y)^4 - 2 (cos x)^2 (cos y)^2 | / | sqrt( x^2 + 2y^2 ) |

subject to
0 <= x <= 10
0 <= y <= 10
x y > 0.75
x + y < 15

:smile:

Scroll to see replies

Reply 1

Fermat

Workin' on it! :smile:

Reply 2

shiny

I knew you couldn't resist - you are a machine!

Reply 3

k@tie

Fermat

Workin' on it! :smile:

aww what a hero!

Reply 4

Fermat

Katie Heskins

aww what a hero!

Uh, some hero. I've got some horrible trig eqns now and I'm stuck!! :frown:

Reply 5

k@tie

Fermat

Uh, some hero. I've got some horrible trig eqns now and I'm stuck!! :frown:

no!! you'll save the day - just you wait and see!

Reply 6

username9816

Poor Fermat, if he is ever stuck then he has nobody to ask on here!

Reply 7

Ralfskini

Here's one for the likes of Fermat..

Our crazy maths teacher walked into the lesson today, asked us to integrate sinxsin2xsin4xsin8xsin16x and then got on with some marking..

Reply 8

makesomenoise

bono

Poor Fermat, if he is ever stuck then he has nobody to ask on here!

How very disrespectful to the other many talented maths students here :rolleyes:

We all help each other out, where one is stuck another may spot the key that unlocks it. It's a great thing, a maths class.

Reply 9

davey_boy

shiny

Find (x, y) that maximises

f(x, y) = | (cos x)^4 + (cos y)^4 - 2 (cos x)^2 (cos y)^2 | / | sqrt( x^2 + 2y^2 ) |

subject to
0 <= x <= 10
0 <= y <= 10
x y > 0.75
x + y < 15

:smile:

Do you have an answer or is it a question you don't know the answer for?

Reply 10

Fermat

davey_boy

Do you have an answer or is it a question you don't know the answer for?

Well, I've got one solution (set) so far. Are these part of the answers?

y=π/2 ±x
======
x and y should be considered subject to those constraints you gave.

Reply 11

shiny

Are you sure those are maxima?

Reply 12

Fermat

shiny

Are you sure those are maxima?

Nope!
I've got the (partial) df/dx=0 and df/dy=0, but, so far the expressions are horrendous and I haven't even considered trying to get d²f/dx², d²f/dy² and d²f/dxdy to determine whether or not they are maxima or minima.
I'm looking to see if I can simplify the original expression first!

Reply 13

Fermat

I don't think I can go any further with this one shiny.

That solution set I gave you before is a local minimum.

I managed to draw the function, and you can see, in the figures attached, that are lots of maxima and minima. But the biggest maximum is at what looks to be (π/2,π/2).

My original solution set was,

y=π/2 + x
and
y=π/2 - x

but as you can see, these give only minima.

Going back to the graphs, then from symmetry there are 4 equivalent maxima.

(π/2, π/2)
(π/2, -π/2)
(-π/2, π/2)
(-π/2, -π/2)

When originally working it out, I got these equations.

cos²x - sin²y = 0
(which gave my (minimas) solution set)

or,

(cos²x - sin²y)[-4sinxcosx(x²+2y²) + x(cos²x - sin²y)] = 0
and,
(cos²x - sin²y)[-4sinycosy(x²+2y²) + 2y(cos²x - sin²y)] = 0

But I couldn't solve these eqns :frown:

So I can't actually prove those results above!

max-and-mins.GIF30.7KB

local-max.gif30.6KB

Reply 14

4Ed

Ralfskini

Here's one for the likes of Fermat..

Our crazy maths teacher walked into the lesson today, asked us to integrate sinxsin2xsin4xsin8xsin16x and then got on with some marking..

can't you just use de Moivre's theorem, substituting expressions for all of those?

Reply 15

Lagz

4Ed

can't you just use de Moivre's theorem, substituting expressions for all of those?

you can just write each sin term in exponential form, simplify it and then perform an easy integration.

Reply 16

Ralfskini

4Ed

can't you just use de Moivre's theorem, substituting expressions for all of those?

yes, maybe but I only do single maths so it took me an hour using p2 trig.

Reply 17

elpaw

does the sqrt term (the denominator) have to be modulused/ated?

Reply 18

shiny

You're allowed to do this problem numerically as long as you specify how.

Rep and SHINY AWARD for the most clever numerical or analytical solutions!

P.S. Yes the denominator should be mod'ed as well.