Respuesta :
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