Find two Real Numbers that differ by 5 and which yield a minimum product.?

Question

find two real numbers that differ by 5 (ex: 1 and 6) which yields a minimum product.

Method: use the vertex form : a(x-p)^2+q

there must be work shown... I'm not allowed to use common sense in this =p

QUOTE:
If you graph it, this will be a parabola, and the vertex will be at the minimum point... but how do you figure out the vertex?

The current unit is parabola and I went up to
P=x(x-5)
=x^2-5x
=(x-5/2)^2-25/4

But I had it all in fractions and it didn't seem right. I knew I was close but I thought it was all wrong. I already forgot what real numbers were as well.

Answer 1

Intuitively I know the answer is:
2.5 * -2.5 = -6.25

But let me show you how we can get the result a little more formally:

Start with the two numbers:
Let x be the smaller number
Let x + 5 be the bigger number

Let f(x) be the product:
f(x) = x(x+5)
f(x) = x² + 5x

If you graph it, this will be a parabola, and the vertex will be at the minimum point... but how do you figure out the vertex?

We need to massage the function so it is in vertex form... so first complete the square. You do this by taking the coefficient on the x-term (5), dividing it in half (2.5) and squaring it (6.25). Now add and subtract this from your equation:

f(x) = x² + 5x + 6.25 - 6.25

The first part is now a perfect square:
f(x) = (x + 2.5)(x + 25) - 6.25

This simplifies to:
f(x) = (x + 2.5)² - 6.25

Compare this to vertex form:
f(x) = a(x - p)² + q

Here's the equation in that form:
f(x) = (x - (-2.5))² + (-6.25)

So:
a = 1
p = -2.5
q = -6.25

If you graph this, you'll find your vertex (minimum point of the parabola) at the point (p, q). This is the point (-2.5, -6.25).

So your two numbers are:
--> x = the smaller number = -2.5
--> x + 5 = the larger number = 2.5

The product of -2.5 and 2.5 is -6.25, which is a minimum.

Answer 2

You are very close to the solution

The current unit is parabola and I went up to
P=x(x-5)
=x^2-5x
=(x-5/2)^2-25/4
you have solved upto this
the above is minimum when (x-5/2)^2 is minimum that is zero

so solution x= 5/2 and minumum value - 25/4

Answer 3

The answer is (rather obviously) 0 and 5. (Product = 0.)

Don't know what work to show. If I asked you for the smallest nonnegative real number, could you show your work to arrive at the answer (0)?

Answer 4

1 and 5...

Answer 5

Que pasa?