For academic math 11 students, this is standard fare. The tutor shows a way to approach it.

Consider the following problem:

Two numbers sum to 20. What is the minimum sum of their squares?

Solution:

Of course we start with let statements:

Let x=one number

Let 20-x be the other number.

Then we call y the sum of their squares and write

y=x^2 + (20-x)^2

This equation represents a quadratic function which can be rendered into vertex form (see my post here):

y=a(x-p)^2+q

From that form, the vertex is given by (p,q).

We need to bring the equation into vertex form. We start by expanding and simplifying.

y=x^2 + (20-x)^2

becomes

y=x^2 + (20-x)(20-x)

which yields, from the foil method,

y=x^2 + 400-20x-20x+x^2

Simplifying, we arrive at

y=2x^2 -40x +400

Now, we begin the change to vertex form – first, by factoring out the coefficient of x^2 from the variable terms only:

y=2(x^2 – 20x)+400

Now, we take half of the coefficient of x^2 (half of -20 is -10), square it (to get 100), then add it inside the brackets. At the same time, to keep the right side equal to what it was before, we subtract twice that amount (or 200) on the outside, because of the factor 2 in front of the brackets:

y=2(x^2 – 20x + 100) + 400 – 200

Now, we realize that x^2-20x+100 is also (x-10)^2. (This can be verified by once again using the foil method.) Therefore,

y=2(x^2 – 20x +100) +200

becomes

y=2(x-10)^2 +200

whose vertex is (10,200). (See my post here about identifying the vertex). From our original setup, 10 represents x, while 200 represents the minumum sum of squares. (If necessary, see the “Let” statements at the beginning to remind yourself.)

If you break 20 into two numbers, then square them and add their squares, you’ll see the smallest sum you can manage is 200: that’s when the numbers are 10 and 10:)

Soon I’ll post a max problem:)

Source:

Travers, Kenneth J. et al. Using Advanced Algebra. Toronto: Doubleday Canada Limited, 1977.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.