Math Problem Statement

If x, y are positive real numbers and 2x+3y=15, find the maximum value of x^{2}y^{3}.

Solution

We are tasked with finding the maximum value of the expression x2y3x^2 y^3 subject to the constraint 2x+3y=152x + 3y = 15, where xx and yy are positive real numbers.

Step 1: Define the Lagrange multiplier function

We can solve this problem using the method of Lagrange multipliers. Define the function we want to maximize as:

f(x,y)=x2y3f(x, y) = x^2 y^3

The constraint is 2x+3y=152x + 3y = 15. Define the constraint function as:

g(x,y)=2x+3y15=0g(x, y) = 2x + 3y - 15 = 0

The Lagrange multiplier method tells us to solve the following system of equations:

f=λg\nabla f = \lambda \nabla g

Where f\nabla f and g\nabla g are the gradients of ff and gg, respectively.

Step 2: Compute the gradients

The gradient of f(x,y)=x2y3f(x, y) = x^2 y^3 is:

f=(fx,fy)=(2xy3,3x2y2)\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( 2x y^3, 3x^2 y^2 \right)

The gradient of g(x,y)=2x+3y15g(x, y) = 2x + 3y - 15 is:

g=(gx,gy)=(2,3)\nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = (2, 3)

Step 3: Set up the system of equations

From the Lagrange multiplier method, we set:

f=λg\nabla f = \lambda \nabla g

This gives us the system:

  1. 2xy3=λ22x y^3 = \lambda \cdot 2 (Equation 1)
  2. 3x2y2=λ33x^2 y^2 = \lambda \cdot 3 (Equation 2)
  3. 2x+3y=152x + 3y = 15 (Constraint)

Step 4: Solve the system of equations

From Equation 1: 2xy3=2λ2x y^3 = 2 \lambda, we can solve for λ\lambda:

λ=xy3\lambda = x y^3

From Equation 2: 3x2y2=3λ3x^2 y^2 = 3 \lambda, substitute λ=xy3\lambda = x y^3 into this:

3x2y2=3(xy3)3x^2 y^2 = 3(x y^3)

Simplify:

x2y2=xy3x^2 y^2 = x y^3

Canceling one xx and one y2y^2 (since x>0x > 0 and y>0y > 0):

x=yx = y

Step 5: Substitute into the constraint

Now substitute x=yx = y into the constraint 2x+3y=152x + 3y = 15:

2x+3x=152x + 3x = 15

5x=155x = 15

x=3x = 3

Since x=yx = y, we also have y=3y = 3.

Step 6: Find the maximum value of x2y3x^2 y^3

Now that we have x=3x = 3 and y=3y = 3, substitute these values into the original expression x2y3x^2 y^3:

x2y3=3233=927=243x^2 y^3 = 3^2 \cdot 3^3 = 9 \cdot 27 = 243

Conclusion:

The maximum value of x2y3x^2 y^3 is 243\boxed{243}.

Would you like further details or have any questions?

Here are five related questions:

  1. What is the method of Lagrange multipliers and when is it used?
  2. How do gradients help in solving optimization problems with constraints?
  3. Can this problem be solved using other methods, such as substitution?
  4. What would happen if the constraint were nonlinear, for example, 2x2+3y2=152x^2 + 3y^2 = 15?
  5. How do the conditions x>0x > 0 and y>0y > 0 influence the solution?

Tip: The method of Lagrange multipliers is particularly useful for optimizing functions subject to linear or nonlinear constraints, as it directly incorporates the constraint into the solution process.

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Math Problem Analysis

Mathematical Concepts

Optimization
Lagrange Multipliers
Calculus
Constraint Optimization

Formulas

f(x, y) = x^2 y^3
g(x, y) = 2x + 3y - 15
Lagrange system: ∇f = λ∇g

Theorems

Lagrange Multiplier Theorem

Suitable Grade Level

Undergraduate Level (Calculus I/II)