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AL2006
11 and 1. 11 times 1 = 11. I double-dare ya to find a different pair with a smaller product.

We want to find two numbers that have a difference of 10, such that its product is minimized.

We will find that the two numbers are 5 and -5, and the minimum possible product is p = -25.

Let's see how to solve this.

So we can define two numbers x and y, such that:

x - y = 10

Now we can write the product of these two numbers as:

p = x*y

From the first equation, we can isolate one of the two variables and write:

x = 10 + y

Now we can replace that in the product equation to get:

p = x*y = (10 + y)*y

p = 10*y + y^2

Now we want to minimize this quadratic equation. Notice that it has a positive leading coefficient, thus, the minimum will be at the vertex of the equation.

Remember that for the general quadratic equation:

Y = a*X^2 + b*X + c

The vertex is at:

X = -b/2a

Then in our case:

p = 10*y + y^2

The vertex is at:

y = -10/2*1 = -5

Then we got the value of y, now we can find the value of x:

x = 10 + y = 10 + (-5) = 5

x = 5

Then the pair of numbers is 5 and -5, and the minimum product is:

p = 5*(-5) = -25

If you want to learn more, you can read:

https://brainly.com/question/3496582