Minimizing the Sums of Squares
Date: 06/12/98 at 16:30:44 From: Anonymous Subject: Minimum and maximum Question: Find two numbers such that their sum is 20 and the sum of their squares is as small as possible. Solution: Let first number be x. Then the second number equals 20 - x. Let y = the sum of their squares. Then: y = x^2 + (20 - x)^2 y = x^2 - 400x - 40x + x^2 y = 2x^2 - 40x + 400 y = 2(x^2 - 20x + 200) y = 2(x - 10)(x - 10) <--- Look below Therefore, x = 10 We know that the answer is 10, but the factor of 2(x^2 - 20x + 200) is not equal to 2(x - 10)(x - 10). So how can we solve this question?
Date: 06/12/98 at 19:08:03 From: Doctor Gary Subject: Re: Minimum and maximum You seem to be confusing the solution of a quadratic equation with finding a minimum value for a quadratic expression. The minimum of a quadratic function in which the coefficient of x^2 is positive is not found at the value of x for which the function itself has a value of zero, but rather at the value of x for which the derivative of the function (the slope of the line tangent to the parabola) is equal to zero. The derivative of 2x^2 - 40x + 400 is 4x - 40. This is equal to zero when x is 10. If you're not yet familiar with derivatives, here's a simpler explanation of why the answer is 10: The two addends we're looking for will have a sum of 10. Consequently, their arithmetic average will be 10. If one of the numbers is n greater than 10, the other number will be n less than 10. The square of the greater number is: (10+n)^2 = 100 + 20n + n^2 The square of the lesser number will be: (10-n)^2 = 100 - 20n + n^2 The sum of the squares will be: 200 + 2n^2 The way to minimize the value of 200 + 2n^2 is to minimize n^2, and that means setting n equal to zero -Doctor Gary, The Math Forum Check out our web site! http://mathforum.org/dr.math//pre
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