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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
Associated Topics:
High School Calculus
High School Number Theory

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