Question

40,6
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S.T the maximum value of x^2y^3z^4 is (a/9)^9 when 2x+3y+4z=a where a greater than 0​

Answer 1

Answer:

the first one is greater

Answer 2

Answer:

The maximum value of x^2y^3z^4 is (a/9) when 2x + 3y + 4z = a, where a > 0.

Step-by-step explanation:

To find the maximum value of x^2y^3z^4, we can use the method of Lagrange multipliers. We want to maximize the function f(x,y,z) = x^2y^3z^4 subject to the constraint g(x,y,z) = 2x + 3y + 4z - a = 0.

First, we set up the Lagrangian function L(x,y,z,λ) = x^2y^3z^4 + λ(2x + 3y + 4z - a).

Next, we take the partial derivatives of L with respect to x, y, z, and λ, and set them equal to zero:

dL/dx = 2xy^3z^4 + 2λ = 0

dL/dy = 3x^2y^2z^4 + 3λ = 0

dL/dz = 4x^2y^3z^3 + 4λ = 0

dL/dλ = 2x + 3y + 4z - a = 0

From the first three equations, we can solve for x, y, z in terms of λ:

x = -λ/y^3z^4

y = -λ/2x^2z^4

z = -λ/3x^2y^3

Substituting these into the fourth equation and simplifying, we get:

λ = (a/9)^(1/9)

Substituting this back into the expressions for x, y, and z, we get:

x = -[(a/9)^(1/9)]/(2y^3z^4)

y = -[(a/9)^(1/9)]/(3x^2z^4)

z = -[(a/9)^(1/9)]/(4x^2y^3)

Finally, we can substitute these expressions for x, y, and z into the original function f(x,y,z) = x^2y^3z^4 to get:

f(x,y,z) = (-[(a/9)^(1/9)]/(2y^3z^4))^2 * (-[(a/9)^(1/9)]/(3x^2z^4))^3 * (-[(a/9)^(1/9)]/(4x^2y^3))^4

Simplifying this expression, we get:

f(x,y,z) = (a/9)

Therefore, the maximum value of x^2y^3z^4 is (a/9) when 2x + 3y + 4z = a, where a > 0.

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