# Finding two unknown numbers

• Jan 2nd 2010, 01:30 PM
brownjovi
Finding two unknown numbers
Determine two numbers whose difference is -4 and whose product is a minimum. Let x be one number and let y be the other.

There is another question similar to this but with maximum instead of minimum, the concept should be the same, right?
• Jan 2nd 2010, 01:37 PM
skeeter
Quote:

Originally Posted by brownjovi
Determine two numbers whose difference is -4 and whose product is a minimum. Let x be one number and let y be the other.

There is another question similar to this but with maximum instead of minimum, the concept should be the same, right?

x - y = -4

x = y - 4

P = xy = (y-4)y

what value of y makes P a minimum?
• Jan 2nd 2010, 06:10 PM
brownjovi
Quote:

Originally Posted by skeeter
x - y = -4

x = y - 4

P = xy = (y-4)y

what value of y makes P a minimum?

Is it two?
• Jan 2nd 2010, 07:19 PM
Sabo
I am a newbie myself, but I think that one way to solve this is to think something along these lines.

If y is negative, the product y(y-4) will be positive.
If y is positive, there will be a few numbers for which the product y(y-4) is negative, and some for which the product will be positive.

In this particular case it is easy to manually figure out for which values of y the product will be negative, since there are only so many positive values of y for which (y-4) will remain negative.
• Jan 2nd 2010, 09:39 PM
Drexel28
If we consider $ax^2+bx+c$ completing the square gives $a\left(x-\tfrac{b}{2a}\right)^2+c-\frac{b^2}{4a}$ which when $a>0$ we may clearly see that $a\left(x-\tfrac{b}{2a}\right)^2+c-\frac{b^2}{4a}\geqslant c-\frac{b^2}{4a}$ so that it attains it's minimum at $x=\frac{b}{2a}$. Geometrically all we did was find the vertex of the parabola.
• Jan 3rd 2010, 01:49 AM
HallsofIvy
Quote:

Originally Posted by Sabo
I am a newbie myself, but I think that one way to solve this is to think something along these lines.

If y is negative, the product y(y-4) will be positive.
If y is positive, there will be a few numbers for which the product y(y-4) is negative, and some for which the product will be positive.

Actually, no. There will be an infinite wet of numbers for which the product y(y- 4) is negative. You cannot assume the number will be a whole number. Completing the square, as Drexel28 suggested, will always give you the minimum or maximum point (vertex of the parabola) of a quadratic function.

Quote:

In this particular case it is easy to manually figure out for which values of y the product will be negative, since there are only so many positive values of y for which (y-4) will remain negative.
Again, no, there are an infinite number of them.
• Jan 3rd 2010, 02:57 AM
Sabo
Right. Stuck in integer land. =/