There are many pairs of numbers (positive and negative) of which the sum is worth the unit. Of those, find the 2 numbers whose sum, double the square of the first number and the square of the 2nd number would give a minimal value.
You keep asking this, but it's still not clear what you want.
What do you mean by "2 numbers whose sum, double the square of the first number and the square of the 2nd number would give a minimal value"?
If the numbers are m and n,
Do you mean that each of m+n and 2m^2 and n^2 are all minimum?
or
m+n + 2m^2 + n^2 is a minimum?
or some other bizarre interpretation?
It's a poorly worded problem.
Also, what is "worth the unit"? Does that mean m+n = 1?
I don't understand the question either, that's why I posted it here and that's exactly how the problem is written word by word, so I have really no idea :(. From what I can interpret, is that we're looking for what sum of two numbers m + n would give a minimal value, now 1 number is 2x(m)^2 and the 2nd number is just at n^2. This is what I think it could mean.
Thank you for your time.
To find the pair of numbers that would give the minimal value, we need to consider all possible pairs of positive and negative numbers whose sum is equal to the unit.
Let's assume the first number in the pair is x, and the second number is y. We want to find the pair (x, y) that minimizes the expression: 2x^2 + y^2.
To solve this problem, we can use optimization techniques. We'll start by considering the possible combinations of x and y, given that their sum is equal to the unit.
1. Generate combinations of x and y:
Since the two numbers' sum is equal to the unit, we have:
x + y = 1
To consider all possible combinations, we'll assign values to x and solve for y:
x = t
y = 1 - t
Here, t is a parameter (ranging from -∞ to +∞) that represents the possible values of x.
2. Substitute the values of x and y into the expression we want to minimize:
Now, we can substitute the values of x = t and y = 1 - t into the expression 2x^2 + y^2:
2t^2 + (1 - t)^2
This expression simplifies to:
2t^2 + (1 - 2t + t^2)
2t^2 - 2t + 1
3. Find the derivative of the expression:
Next, we'll find the derivative of the expression with respect to t, and set it equal to zero to find the critical points:
d/dt (2t^2 - 2t + 1) = 0
4. Solve for critical points:
Solving the derivative equal to zero gives us:
4t - 2 = 0
t = 1/2
5. Find the minimal value:
Now, we'll substitute the value of t = 1/2 back into the expression 2t^2 - 2t + 1 to find the minimal value:
2(1/2)^2 - 2(1/2) + 1
1/2
Therefore, the minimal value is 1/2.
To summarize, to find the pair of numbers that would give the minimal value of 2x^2 + y^2, we need to solve the equation x + y = 1 and substitute the values of x and y into the expression. By finding the critical points and evaluating the expression at those points, we can determine the minimal value. In this case, the minimal value is 1/2.