If the sum of two positive real numbers is 1 then what is the minimum ...
Let the two numbers be a and b.
a + b = 1
Maximum value of ab will occur at a = b = 0.5
And the maximum value of ab = 0.5 x 0.5 = 0.25
Maximum value of 2ab = 2 * 0.25 = 0.5
Now, (a + b)2 = 1
a2 + b2 + 2ab = 1
a2 + b2 = 1 - 2ab
a2 + b2 will be minimum when 2ab will be maximum i.e. 0.5.
Minimum value of a2 + b2 = 0.5
Answer: 0.5
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If the sum of two positive real numbers is 1 then what is the minimum ...
The minimum value of the sum of the squares of two positive real numbers can be found using the concept of optimization.
Problem:
Let's assume we have two positive real numbers, x and y, with the sum of 1. We need to find the minimum value of the sum of their squares, x^2 + y^2.
Approach:
To solve this problem, we can use the concept of optimization by considering the given condition and trying to minimize the sum of squares.
Step 1: Formulating the problem mathematically
Let's formulate the problem mathematically:
We need to minimize f(x, y) = x^2 + y^2, given the constraint x + y = 1.
Step 2: Solving the problem
We can solve this problem using the method of Lagrange multipliers, which helps us find the minimum or maximum of a function subject to constraints.
Step 2.1: Formulating the Lagrangian function
The Lagrangian function is given by:
L(x, y, λ) = f(x, y) - λ(g(x, y)), where g(x, y) represents the constraint equation and λ is the Lagrange multiplier.
In our case, the Lagrangian function becomes:
L(x, y, λ) = x^2 + y^2 - λ(x + y - 1)
Step 2.2: Finding the partial derivatives
Next, we find the partial derivatives of the Lagrangian function with respect to x, y, and λ:
∂L/∂x = 2x - λ
∂L/∂y = 2y - λ
∂L/∂λ = -(x + y - 1)
Step 2.3: Setting the partial derivatives to zero
To find the critical points, we set the partial derivatives to zero:
2x - λ = 0
2y - λ = 0
-(x + y - 1) = 0
Solving these equations simultaneously, we get:
x = y = 1/2
λ = 2
Step 2.4: Checking the nature of the critical point
To determine if the critical point is a minimum, maximum, or a saddle point, we need to analyze the second-order partial derivatives. However, in this case, we can observe that the function x^2 + y^2 is always positive or zero. Therefore, the minimum value of x^2 + y^2 occurs when x = y = 1/2.
Step 3: Substituting the values back
Substituting x = y = 1/2 in the function f(x, y) = x^2 + y^2, we get:
f(1/2, 1/2) = (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2
Hence, the minimum value of the sum of the squares of two positive real numbers, given their sum is 1, is 1/2 or 0.5.
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