Key Concepts and Formulas

• Equation of a Line: Given two points $(x_1, y_1)$ and $(x_2, y_2)$, the equation of the line passing through them is given by $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$.
• Lagrange Multipliers (Optimization with Constraint): To find the maximum or minimum value of a function $f(x, y)$ subject to a constraint $g(x, y) = c$, we can use the method of Lagrange multipliers. We introduce a Lagrange multiplier $\lambda$ and solve the following system of equations:
  • $\nabla f(x, y) = \lambda \nabla g(x, y)$
  • $g(x, y) = c$
• Gradient: The gradient of a function $f(x, y)$ is given by $\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$.

Step-by-Step Solution

Step 1: Find the equation of the line.

The line passes through the points $(50 + \alpha, 0)$ and $(0, 50 + \alpha)$. We use the two-point form of the equation of a line: $\frac{y - 0}{x - (50 + \alpha)} = \frac{50 + \alpha - 0}{0 - (50 + \alpha)}$ $\frac{y}{x - (50 + \alpha)} = \frac{50 + \alpha}{-(50 + \alpha)} = -1$ $y = -x + 50 + \alpha$ $x + y = 50 + \alpha$ This is our constraint equation.

Step 2: Set up the Lagrangian function.

We want to maximize $f(x, y) = xy^4$ subject to the constraint $g(x, y) = x + y = 50 + \alpha$. We define the Lagrangian function as: $L(x, y, \lambda) = xy^4 - \lambda(x + y - (50 + \alpha))$

Step 3: Find the partial derivatives and set them to zero.

We need to find the partial derivatives of $L$ with respect to $x$, $y$, and $\lambda$ and set them equal to zero:

• $\frac{\partial L}{\partial x} = y^4 - \lambda = 0 \implies \lambda = y^4$ ...(1)
• $\frac{\partial L}{\partial y} = 4xy^3 - \lambda = 0 \implies \lambda = 4xy^3$ ...(2)
• $\frac{\partial L}{\partial \lambda} = -(x + y - (50 + \alpha)) = 0 \implies x + y = 50 + \alpha$ ...(3)

Step 4: Solve the system of equations.

From equations (1) and (2), we have: $y^4 = 4xy^3$ Since $y \neq 0$ (otherwise, $xy^4 = 0$, which is not a maximum), we can divide both sides by $y^3$: $y = 4x$

Now, substitute $y = 4x$ into equation (3): $x + 4x = 50 + \alpha$ $5x = 50 + \alpha$ $x = \frac{50 + \alpha}{5}$

Then, $y = 4x = 4\left(\frac{50 + \alpha}{5}\right) = \frac{4(50 + \alpha)}{5}$

Step 5: Verify the solution.

The point $(x, y) = \left(\frac{50 + \alpha}{5}, \frac{4(50 + \alpha)}{5}\right)$ lies on the line $y = 4x$.

Common Mistakes & Tips

• Remember to consider the constraint equation when optimizing a function.
• When using Lagrange multipliers, make sure to find all partial derivatives and solve the resulting system of equations.
• Be careful when dividing by variables. Ensure they are not zero.

Summary

We are given the objective function $xy^4$ and the constraint $x+y=50+\alpha$. Using the method of Lagrange multipliers, we set up the Lagrangian function and found the partial derivatives with respect to $x$, $y$, and $\lambda$. Setting these derivatives to zero, we solved the system of equations to find the relationship $y=4x$. Therefore, the point $(x, y)$ lies on the line $y = 4x$.

Final Answer

The final answer is \boxed{y = 4x}, which corresponds to option (A).