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Topic: "Without the help of calculus ..."
Replies: 13   Last Post: Jan 9, 2014 8:04 AM

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 @less@ndro Posts: 221 Registered: 12/13/04
Re: "Without the help of calculus ..."
Posted: Dec 27, 2013 11:31 PM

> 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+y-6)^2 + (x-7/2)^2 -213/4

This gives 7x + 4y = 213/4 - (x+y-6)^2 + (x-7/2)^2 <= 213/4

and the value 213/4 is assumed at the solution (x,y) = (7/2,5/2).

--
Marc

Date Subject Author
12/25/13 Port563
12/25/13 Bart Goddard
12/25/13 Port563
12/26/13 Bart Goddard
12/26/13 Port563
12/26/13 quasi
12/26/13 Port563
12/26/13 David Bernier
12/27/13 David Bernier
12/26/13 Leon Aigret
12/26/13 Port563
12/26/13 Leon Aigret
12/27/13 @less@ndro
1/9/14 Bill Taylor