The minimum value of 3x - 4y subject to the condition x^2y^3 = 6 can be found using the method of Lagrange multipliers.

Let f(x, y) = 3x - 4y be the objective function and g(x, y) = x^2y^3 - 6 = 0 be the constraint function.

We want to find the values of x and y that minimize f(x, y) while satisfying the constraint g(x, y) = 0.

The Lagrangian function is defined as L(x, y, λ) = f(x, y) - λg(x, y), where λ is the Lagrange multiplier.

Taking partial derivatives and setting them equal to zero, we have:

∂L/∂x = 3 - 2λxy^3 = 0
∂L/∂y = -4 - 3λx^2y^2 = 0
∂L/∂λ = -x^2y^3 + 6 = 0

From the first equation, we have 2λxy^3 = 3. Rearranging, we get λ = 3/(2xy^3).

Substituting this value of λ into the second equation, we have -4 - 3(3/(2xy^3))x^2y^2 = 0.

Simplifying, we have -4 - 9x^2y^2/(2xy^3) = 0. Canceling out common factors, we get -4 - 9x/(2y) = 0.

Solving for x, we have x = -8y/9.

Substituting this value of x into the constraint equation, we have (-8y/9)^2y^3 - 6 = 0.

Simplifying, we have 64y^6/81 - 6 = 0.

Multiplying through by 81, we have 64y^6 - 486 = 0.

Solving for y, we have y^6 = 486/64 = 27/4.

Taking the sixth root of both sides, we have y = (27/4)^(1/6).

Substituting this value of y into the equation x = -8y/9, we have x = -8(27/4)^(1/6)/9.

Therefore, the minimum value of 3x - 4y subject to the condition x^2y^3 = 6 is given by substituting the values of x and y into the objective function:

f(x, y) = 3x - 4y = 3(-8(27/4)^(1/6)/9) - 4(27/4)^(1/6).

Simplifying this expression will give the minimum value.