Verify that the sum of three quantities x, y, and z, whose product is a constant k, is maximum when these three quantities are equal.

x y z = k

sum = s = x + y + z

ds = ds/dx dx + ds/dy dy + ds/dz dz
= 0 for max or min
but ds/dx = ds/dy = ds/dz = 1
so
ds = dx + dy + dz = 0
so max when dx = -dy - dz
but
x = k/(yz)
dx/dy = -kz /(yz)^2
dx/dz = -ky /(yz)^2
so
dx = -[-kz/((yz)^2 ]dy - [-ky/(yz)^2]dz
so k y = k z
so y = z

xyz=k therefore z=k/xy

S=x+y+k/xy
dS=(δS/δx)dx+(δS/δy)dy
=(1-k/x^2y)dx+(1-k/y^2x)dy
For min or max
1-k/x^2y=0,1-k/y^2x=0
Therefore k/x^2y=1 so x=k/xy
And k/y^2x=1 so y=k/xy
So x=y=z=k/xy

To verify that the sum of three quantities x, y, and z, whose product is a constant k, is maximum when these three quantities are equal, we can use the method of optimization.

Let's assume x, y, and z are the three quantities, and their product is k:

x * y * z = k

To find the maximum sum of x, y, and z, we can use the AM-GM inequality. According to the AM-GM inequality:

(x + y + z) / 3 ≥ (xyz)^(1/3)

Since the product xyz is constant and equal to k, we can rewrite the inequality as:

(x + y + z) / 3 ≥ k^(1/3)

Now, to find the maximum value of x + y + z, we need to minimize the difference between the left-hand side of the inequality and k^(1/3). This occurs when x = y = z.

Let's substitute x = y = z into the inequality:

(3x) / 3 ≥ k^(1/3)

x ≥ k^(1/3)

Since x, y, and z are equal, we can substitute any variable with x, which gives us:

x + y + z = 3x ≥ 3k^(1/3)

Therefore, the sum of three quantities x, y, and z is maximum when these three quantities are equal.

Note: It's important to keep in mind that this proof assumes a positive constant product (k > 0) and that x, y, and z are positive real numbers.

To verify that the sum of three quantities x, y, and z, whose product is a constant k, is maximum when these three quantities are equal, we can use the method of optimization.

Let's assume that x, y, and z are the three quantities, and their product is k. Mathematically, we have:

xy * z = k

To prove that their sum is maximum when x = y = z, we can use algebraic manipulation and the concept of inequality.

We start by manipulating the equation xy * z = k as follows:

xy * z = k
xy = k / z

Now, we can express the sum of x, y, and z in terms of the variables x and z:

x + y + z = x + x * (k / (xz)) + z
= x * (1 + k / (xz)) + z

To find the maximum value of this expression, we need to take the derivative with respect to x and set it equal to zero. Differentiating the above expression with respect to x, we get:

d/dx (x * (1 + k / (xz)) + z) = (1 + k / (xz)) - (k / (xz^2))
= 1/x - k / (xz^2)

Setting the derivative equal to zero and solving for x, we can simplify the equation as follows:

1/x - k / (xz^2) = 0
1 - k / (xz) = 0
xz = k

From the equation xz = k, we can conclude that x = k/z. Substituting this value of x back into the expression for the sum of x, y, and z, we have:

x + y + z = (k/z) * (1 + k / ((k/z)z)) + z
= (k/z) * (1 + z/z) + z
= (k/z) * 2 + z
= 2k/z + z

To maximize the sum x + y + z, we need to find the value of z that maximizes this expression. Taking the derivative with respect to z and setting it equal to zero, we have:

d/dz (2k/z + z) = -2k/z^2 + 1 = 0

Simplifying this equation and solving for z, we get:

-2k/z^2 + 1 = 0
2k = z^2
z^2 = 2k
z = sqrt(2k)

Therefore, the maximum value of the sum x + y + z occurs when z = sqrt(2k). Since x = k/z, substituting this value of z, we have x = k/sqrt(2k) = (sqrt(2k))/2. Similarly, y will also be (sqrt(2k))/2.

Thus, the values of x, y, and z that maximize their sum are x = (sqrt(2k))/2, y = (sqrt(2k))/2, and z = sqrt(2k). In other words, the sum of x, y, and z is maximum when x = y = z.

By following this process of optimization, we have mathematically verified that the sum of three quantities x, y, and z, whose product is a constant k, is maximum when these three quantities are equal.