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 May 30th, 2009, 06:10 AM #1 Member   Joined: Jun 2008 Posts: 45 Thanks: 0 largest possible value Hi! What is the largest possible value of $\sqrt{(ax+by)^2 + (cx+dy)^2}$ when $x^2+y^2=1$? It seems that there is more clever way to solve the problem than using this standard method from mathematical analysis, besides I don't have to find x and y when maximum is reached. I definitely think I must use something of $\frac{n}{\frac{1}{a_1} + \ldots + \frac{1}{a_n}} \leq \sqrt[n]{a_1 \ldots a_n} \leq \frac{a_1 + \ldots + a_n}{n} \leq sqrt{\frac{a_1^2 + \ldots + a_n^2}{n}}$ or $(a_1x_1 + \ldots + a_nx_n)^2 \leq (a_1^2 + \ldots + a_n^2)(x_1^2 + \ldots + x_n^2)$, but I can't get on with mine, Mathematica and Maple results. For example, taking a=1, b=2, c=3, d=4, I get $\sqrt{30}$, but both Mathematica and Maple give $\sqrt{15 + \sqrt{221}}$. Okay, $\sqrt{15 + \sqrt{221}}$ seems to be correct, but how can get it?
 May 30th, 2009, 10:35 AM #2 Senior Member   Joined: Nov 2007 Posts: 258 Thanks: 0 Re: largest possible value It's equivalent to maximizing $f(x,y)= (ax+by)^2+(cx+dy)^2$. The two gradients will be colinear: $\bigtriangledown f(x,y)= (2a(ax+by) + 2c(cx+dy), 2b(ax+by) + 2d(cx+dy) = k(2x, 2y)$ $\bigtriangledown f(x,y)= (a(ax+by) + c(cx+dy), b(ax+by) + d(cx+dy) = k(x, y)$ $x(a^2+c^2)+y(ab+cd)= kx$ $y(b^2+d^2)+x(ab+cd)= ky$ $(a^2+c^2)+y/x(ab+cd)= k$ $(b^2+d^2)+x/y(ab+cd)= k$ $(a^2+c^2)+y/x(ab+cd)= (b^2+d^2)+x/y(ab+cd)$ $\frac{(a^2+c^2) - (b^2+d^2)}{(ab+cd)}= (x/y-y/x)$ $\frac{(a^2+c^2) - (b^2+d^2)}{(ab+cd)}= (x^2-y^2)/(xy)$ $\frac{(a^2+c^2) - (b^2+d^2)}{(ab+cd)}= (1-2y^2)/(\sqrt{1-y^2}y)$ Set $m=\frac{(a^2+c^2) - (b^2+d^2)}{(ab+cd)}$ : $m(\sqrt{1-y^2}y)= (1-2y^2)$ $m^2(y^2-y^4)= (1-2y^2)$ $y^4m^2 - y^2(m^2+2) + 1= 0$ Now you have a quadratic in y^2 . Not too clean but it should work. Hope this helps! Don't forget that I disregarded the initial square root.
 June 4th, 2009, 01:59 PM #3 Member   Joined: Jun 2008 Posts: 45 Thanks: 0 Re: largest possible value Thank you brunojoyal very much. I will try to enter into this.
 June 6th, 2009, 03:06 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2009 The expression = ?((a² + b² + c² + d² + (b² + d² - a² - c²)cos(2t) + (2ab + 2cd)sin(2t))/2), where x = sin(t) and y = cos(t). Let r = ?((b² + d² - a² - c²)² + (2ab + 2cd)²). The expression has maximum value ?((a² + b² + c² + d²)/2 + r/2). (Do you see why?)
 September 5th, 2015, 06:09 AM #5 Senior Member   Joined: Sep 2015 From: 4th Dimension Posts: 146 Thanks: 13 Math Focus: Everything (a little bit) I still feel there shud be a direct method ( clever one)

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