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copied!<p>Lets say you have two numbers <code>x</code> and <code>y</code> such that <code>x*y = z</code> (where <code>z</code> is your constant number)</p> <ul> <li><p>We want <code>xy =z</code> for a constant <code>z</code> and we want to minimize <code>f(x) = x+y</code></p></li> <li><p>We know that <code>xy = z</code> <code>so x = z/y</code> (assuming z is not 0, otherwise it's fairly easy to prove there is no solution)</p></li> <li><p>So we need to minimize <code>f(x) = z/x + x</code></p></li> <li><p>Derivate <code>f(x)</code> to get <code>f'(x) = 1 - z/(x^2)</code></p></li> <li><p>Find min/max points by equating the derivative to 0 (using <a href="http://en.wikipedia.org/wiki/Fermat%27s_theorem_%28stationary_points%29" rel="nofollow">Fermat's theorem</a> getting:</p> <p><code>z/(x^2) = 1</code> which brings you <code>x^2 = z</code></p></li> </ul> <p>Meaning the only possible point of maximum is <code>x^2 = z</code> which means <code>x=√z</code> which means <code>y=x</code></p> <p><strong>However, you already knew that.</strong></p> <ul> <li><p>We know that for every <code>x</code> and <code>y</code> it holds that <code>(x+y)(x-y) = x^2-y^2</code></p></li> <li><p>If we write <code>x=a+b</code> and <code>y=a-b</code> for some a and b we get that <code>(a-b)(a-b)</code> is maximal if y is 0 meaning in our example <code>x=y</code>.</p></li> <li><p>Which bring us to the strategy Gilbert Le Blanc suggested if you're only accepting integer solutions, start at the root of <code>z</code> and try increasing decreasing from it. The first number you match is the result.</p></li> </ul>

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