find the maximum and minimum values of x+y+z subject to 1\x+1\y+1\z=1
Answers
Answer:
the minimum value of x + y + z subject to the constraint 1/x + 1/y + 1/z = 1 is 9, which occurs when x = y = z = 3.
Step-by-step explanation:
We can use the method of Lagrange multipliers to find the maximum and minimum values of x + y + z subject to the constraint 1/x + 1/y + 1/z = 1.
Let f(x, y, z) = x + y + z and g(x, y, z) = 1/x + 1/y + 1/z - 1. We want to find the maximum and minimum values of f subject to the constraint g = 0.
The Lagrange function is L(x, y, z, λ) = f(x, y, z) - λg(x, y, z) = x + y + z - λ(1/x + 1/y + 1/z - 1).
Taking the partial derivatives of L with respect to x, y, z, and λ, we get:
∂L/∂x = 1 + λ/x^2 = 0
∂L/∂y = 1 + λ/y^2 = 0
∂L/∂z = 1 + λ/z^2 = 0
∂L/∂λ = 1/x + 1/y + 1/z - 1 = 0
Solving for x, y, z, and λ, we get:
x = y = z = (3/λ)^(1/3)
1/x + 1/y + 1/z = 3/x = 3(λ/3)^(1/3) = 1
λ = 3^(2/3)
Substituting back into f(x, y, z), we get:
f(x, y, z) = 3(3/λ)^(1/3) = 3^(2/3)
Therefore, the maximum value of x + y + z subject to the constraint 1/x + 1/y + 1/z = 1 is 3^(2/3).
To find the minimum value, we can use the AM-HM inequality, which states that for any positive numbers a, b, and c:
(a + b + c)/3 ≥ 3/(1/a + 1/b + 1/c)
Applying this inequality to x, y, and z, we get:
(x + y + z)/3 ≥ 3/(1/x + 1/y + 1/z) = 3
Multiplying both sides by 3, we get:
x + y + z ≥ 9
Therefore, the minimum value of x + y + z subject to the constraint 1/x + 1/y + 1/z = 1 is 9, which occurs when x = y = z = 3.
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