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.
To find the pair of numbers that satisfies the given conditions and gives a minimal value, we can express the problem mathematically and use some basic calculus concepts.
Let's denote the two numbers as x and y. According to the given conditions, we have:
x + y = 1 (Equation 1)
2x^2 + y^2 = min (Equation 2)
To find the minimal value of the equation (Equation 2), we need to minimize it using calculus. First, let's solve Equation 1 for y:
y = 1 - x
Now, substitute the value of y in Equation 2:
2x^2 + (1 - x)^2 = min
Expand the equation:
2x^2 + 1 - 2x + x^2 = min
Combine like terms:
3x^2 - 2x + 1 = min
To find the minimum value, we can take the derivative of this equation with respect to x and set it equal to zero:
d(3x^2 - 2x + 1)/dx = 0
6x - 2 = 0
Solve for x:
6x = 2
x = 2/6
x = 1/3
Now, substitute the value of x back into Equation 1 to find y:
y = 1 - x
y = 1 - 1/3
y = 2/3
So, the pair of numbers that satisfies the given conditions and gives a minimal value are (1/3, 2/3).
To verify that this pair is indeed the minimum, you can substitute the values of x and y back into Equation 2 and compare the value with other potential pairs.