
Re: "Without the help of calculus ..."
Posted:
Dec 27, 2013 11:31 PM


Port563 <reader80@eternalseptember.org> wrote: > This easy problem was very recently presented to me. > > Without any recourse to calculus, establish maximum and minimum values for: > > 7x + 4y > > where 2x^2 + y^2 + 2xy  12x 8y 5 = 0, x and y being real. > > I found cheapo ways, e.g. > > * relying on inspection, then deducing the limits (tangents) of a set of > "1.75 sloped" secants to an ellipse, etc.; > > alternatively > > * rejigging the expression's terms so it appears as a quadratic in y, > finding the discriminant and then simple manipulation to find/prove extrema; > > but, given the source of the problem, there might also be an elegant route I > missed. > > Anyone?
I do not know if it is elegant, but at least it does only need completing squares and no further calculus: try to rewrite the above equation 2x^2 + y^2 + 2xy  12x 8y 5 = 0 as k(7x+4y) + S + r = 0 where k is a suitable nonzero integer, S is a sum of squares and r is a real number.
For the minimum value one can take:
2x^2 + y^2 + 2xy  12x  8y  5 = 2(7x+4y) + (x+y)^2 + (x+1)^2  6
This gives 7x + 4y = ((x+y)^2 + (x+1)^2)/2  3 >= 3 and the value 3 is assumed at the solution (x,y) = (1,1).
For the maximum value one can take:
2x^2 + y^2 + 2xy  12x  8y  5 = 7x+4y + (x+y6)^2 + (x7/2)^2 213/4
This gives 7x + 4y = 213/4  (x+y6)^2 + (x7/2)^2 <= 213/4
and the value 213/4 is assumed at the solution (x,y) = (7/2,5/2).
 Marc

