**Q:** Two numbers hava a difference of 16. Find the numbers if their product is a minimum.

**A:** Let x = first number

Let y = second number

Equation 1: x-y = 16

Equation 2: xy=P , where P is the product.

You need to isolate for x or y in Equation 1, and then substitute it into Equation 2. You always want to work with the equation that has the quantity you are trying to maximize or minimize.

Isolating in Equation 1:

x = 16+y

Substitute this into Equation 2:

P = xy

= (16+y)y

= 16y + y^2 [note that y^2 means y squared]

You now have to turn this into a parabola form by completing the square. [Take half of 16, then square it to get 64. Add this to get a perfect square, then subtract it so that you are only adding zero to the equation.]

= y^2 + 16y

= y^2 + 16y + 64 – 64

= (y^2 +16y + 64) – 64

P = (y+8)^2 – 64

This is the equation of a parabola, in this case, P is acting like y, and y is acting like x. (let me know if you’re confused by this!)

The vertex is at (-8, -64). A graph of it is shown here:

Notice that the minimum is -64 and it occurs when y = -8.

The only thing left to do is find x, the other number:

x = 16 + y

x = 16 + -8

x = 8

So the two numbers are 8 and – 8. They have a difference of 16 and their product is a minimum, as displayed in the parabola.

Hope this helps!

## Follow Us!