Answer: {-5, 5}. Product is -25, which is the minimum.
Explanation:
let a, b denote the two numbers. We know that b-a=10.
We are looking for a minimum over the product a*b.
[tex]ab=a(a+10)= a^2+10a[/tex]
One can minimize this using derivatives. In case you have not yet had derivatives, you can also use the vertex of a parabola (since the above is a quadratic form):
[tex]a^2+10a= (a+5)^2-25\implies x_{min}=-5[/tex]
The minimum is at the vertex a=-5 and so b=5
Their distance is 10, and their product attains the minimum value of all possiblities -25.
We want to find the minimum for a product between two numbers for a given restriction.
The two numbers we want to get are 5 and -5, and the minimum of the product is P = -25.
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Let's see how to solve this.
We have the restriction:
"the two numbers have a difference of 10"
So if the two numbers are A and B, we have that:
A - B = 10
Now we want to minimize their product.
The product is written as:
P = A*B
Using the restriction, we can write:
A = 10 + B
Then we get:
P = (10 + B)*B = B^2 + 10*B
This is a quadratic equation with a positive leading coefficient, thus, graph of the parabola opens up.
This means that the minimum of P is at the vertex of the parabola.
Remember that for a general parabola:
y = a*x^2 + b*x + c
The x-value of the vertex is given by:
x = -b/2a
Then for our equation:
P = B^2 + 10*B
The vertex is at:
B = -10/2*1 = -5
Now with the equation:
A = 10 + B
We can find the value of A.
A = 10 + (-5) = 5
Then we have:
The value of the product is P = 5*(-5) = -25.
The minimum of the product is -25.
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