<p>maybells:</p>
<p>As the problem is stated, it cannot be solved. There are infinitely many real numbers x,y such that x + y = 18. If, however, the problem aks you to <em>maximize</em> the product of the numbers, then it is very solvable.</p>
<p>Let's generalize and consider numbers that add to be S. In your problem, S = 18. If one number is x, then the other is (S - x). Thus, the product is</p>
<p>x(S - x) = Sx - x^2.</p>
<p>The problem then resolves itself into one of maximizing the quadratic f(x) = -x^2 + Sx. The quadratic term is negative, so therefore this one opens downward and the global maximum occurs at the vertex. There are two ways to find the vertex:</p>
<p>1) Set the derivative equal to zero and solve for x. In this case:</p>
<p>f'(x) = -2x + S
-2x + S = 0
-2x = -S
x = S/2</p>
<p>2) Remember accurately the formula from Algebra II stating that the x-coordinate of the vertex of ax^2 + bx + c is -b/2a. In this case, a = -1, b = S and c = 0; therefore, the x-coordinate of the vertex is</p>
<p>-b/2a = -S/[ 2(-1) ] = -S/-2 = S/2</p>
<p>So, if one number is S/2, then the other number is</p>
<p>S - S/2 = 2S/2 - S/2 = S/2.</p>
<p>Thus, the maximum product occurs when both numbers are exactly half of the target sum. In this case, S = 18, meaning both numbers should be 18/2 = 9.</p>
<p>hope this helps,
nilkn</p>