All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Maximizing product (Posted on 2013-09-19)
In a triangle ABC two items are defined : the side AB=c and the height from point C to this side Hc= k.

What is the maximum value of the product Ha*Hb* Hc ?

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Final Correction !! Comment 4 of 4 |
I have mistakenly substituted Hc as c instead of k, which messed up both the original solution as well as the correction, so here is the full final corrected solution :-

As Hc is a given parameter (Hc=k), we need only to maximize

Ha*Hb.

We draw a triangle ABC, where a=b. We draw through point C, a

parallel to side c. The distance between the parallels is Hc=k.

Shifting the point C along the parallel represents all the

possible triangles having the given parameters.

We now have to find the distance f to which point C has to be

shifted so as to result in the asked for  maximization of

Ha*Hb*Hc.

Denoting : e=c/2, we have the area common to all shifted triangles

as S=e*k.

Denoting the angle of point C as Gamma , we express the triangle

area also as :
S=e*k=a*b*Sin(Gamma)/2

Sin(Gamma)=2*e*k/a*b

a*b = 2*S/Sin(Gamma)

Now, Ha=2*S/a and Hb=2*S/b ,and

Ha*Hb = 4*S^2/(a*b)

substituting the above expression for a*b, we get :

Ha*Hb = 2*S*Sin(Gamma)=c*k*Sin(Gamma)

and multiplying by Hc=k, we get the expression we need to

maximize :       Ha*Hb*Hc = c*k^2*Sin(Gamma)

as it is seen, we need to maximize Sin(Gamma)

we now express Sin(Gamma) as a function of the shift f and will

find the f which maximizes Sin(Gamma) :

For a shifted triangle we get :

Sin(Gamma) = 2*e*k/{sqrt[k^2+(e-f)^2]*sqrt[k^2+(e+f)^2]}

and we need to minimize the denominator of this expression.

After some Algebra we reach the following :

(Denominator)^2 = (k^2+e^2)^2 + f^2*[2*(k^2 - e^2) + f^2]

We distinguish now 2 cases :

1.    k>e

minimization of the denominator clearly requires  f=0

resulting in the case :  a=b , and

Sin(Gamma) = 2*e*k/(k^2 + e^2)

and

Ha*Hb*Hc = c*k^2*Sin(Gamma) = c^2*k^3/(k^2+e^2) =

4*c^2*k^3/(4*k^2+c^2)

2.   k<e

We now reorganize the expression to be minimized, to the

following form :

F=  (Denominator)^2 = k^4+2*k^2*(e^2+f^2)+(e^2-f^2)^2

and differentiate with respect to f  for finding the minimum :

dF/d(f^2) = 2*k^2-2*e^2+2*f^2 = 0

f^2 = e^2 - k^2

substituting into the expression for Sin(Gamma) we get :

Sin(Gamma) = 2*e*k/(2*e*k) = 1 ,

and -

Ha*Hb*Hc  =  k*c^2

 Posted by Dan Rosen on 2013-09-25 11:54:07

 Search: Search body:
Forums (0)