Question

find the maximum and minimum values of x+y+z subjected to 1/x+1/y+1/z=1​

Answer 1

Answer:

Another perspective on this question is that the equation x + y + z =1 defines a plane in the three dimensional space. Of interest is the portion of the this plane lying in the first octant. This portion of the plane is an equilateral tringle. The maximum of xyz lies at the center of this equilateral triangle. By analogy to linear programming, the minima will lie at the corners. These are the points (1,0,0), (0,1,0), (0,0,1). This reasoning is probably 

Answer 2

To find the maximum and minimum values of x+y+z subjected to 1/x+1/y+1/z=1, we can use the method of Lagrange multipliers.

Let f(x,y,z) = x+y+z and g(x,y,z) = 1/x+1/y+1/z-1.

We need to find the critical points of f subject to the constraint g=0. Using Lagrange multipliers, we get the following system of equations:

∇f = λ∇g

g = 0

Solving these equations, we get the critical point (x,y,z) = (a,a,a), where a = 3^(1/3).

To determine if this point is a maximum or minimum, we need to consider the second partial derivatives of f. We find that f_xx = f_yy = f_zz = 0, and f_xy = f_xz = f_yz = 1. Therefore, the Hessian matrix is the 3x3 identity matrix, which is positive definite. Hence, the critical point (a,a,a) is a minimum for f subject to the constraint g=0.

• Therefore, the minimum value of x+y+z is $3^(4/3)$, which is obtained when $x=y=z=3^(1/3).$

• To find the maximum value, note that as x, y, and z approach infinity, 1/x+1/y+1/z approaches zero. Thus, there is no maximum value for x+y+z subject to 1/x+1/y+1/z=1.

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