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Find the possible values of x,y,z at which f(x,y,z)=4x²+4y²+3z²−3x−2y−z assumes its minimum value.

User Masiar
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1 Answer

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Answer: Therefore at (x, y, z) = (3/8, 1/4, 1/6), the function f(x, y, z) assumes its minimum values.

Explanation:

First, let's find the partial derivatives of f(x, y, z) with respect to x, y, and z:

∂f/∂x = 8x - 3

∂f/∂y = 8y - 2

∂f/∂z = 6z - 1

To find the critical points, we set these partial derivatives equal to zero and solve for x, y, and z:

8x - 3 = 0 --> x = 3/8

8y - 2 = 0 --> y = 1/4

6z - 1 = 0 --> z = 1/6

Therefore, the critical point is (x, y, z) = (3/8, 1/4, 1/6).

To determine if this point is a minimum, we can use the second derivative test.

∂^2f/∂x^2 = 8

∂^2f/∂y^2 = 8

∂^2f/∂z^2 = 6

Since the second partial derivatives are all positive, the critical point (3/8, 1/4, 1/6) is a local minimum.

User Aditya Kadakia
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