Implicit Diff Q (1 Viewer)

[azureus88]

Prove that a cyclic quadrilateral has the maximum area of all quadrilaterals with the same side lengths in hte same order. (by letting sides be a,b,c,d and letting P be the included angle of a,b and Q be included angle of c,d)

I got up to the part where:

dA/dP = absin(P+Q)/2sinQ = 0
P+Q = pi since 0<P<pi and 0<Q<pi

but how do i make sure this is the maximum point and not minimum without resorting to common sense. Second derivative seems to tedious and i dont think you can test changes in sign around P as there are two variables.

 

[cyl123]

Firstly, you have to consider cases where the quadrilateral is an arrowhead, ie one of the angles is > pi. But it is physically obvious that but just inverting the angle greater than pi, you will increase the area. ie. the quad with greatest area will have all angles less than pi.

i suppose you can test P = pi - Q +1 and P = pi - Q -1 while assuming Q remains constant as P changes (THIS DOESNT MEAN THAT Q IS COMPLETELY CONSTANT, you are only seeing if given Q, what P will maximise area) and is an angle greater than 0 and less than pi (or else P is less or eqaul to 0) This gives:
at p = pi -Q +1
dA/dP = absin(pi+1)/sinQ < 0 as absin(pi+1)<0 and sinQ>0
at p = pi -Q -1
dA/dP = absin(pi-1)/sinQ > 0 as absin(pi-1)>0 and sinQ>0.

Thus proving P= pi - Q +1 maximises angle.

(I suppose a more tedious version is to use the intrinsic restriction on/relation between P and Q which is c^2 + d^2 - 2cdcosQ = a^2 +b^2 -2abcosP and substitute out Q with the relation and differentiate in terms of one variable only)

 

[azureus88]

ok thanks, i get it now