To find the Lagrange multiplier associated with the equality constraint for the given utility maximization problem, follow these steps: 1. **Set up the Lagrangian**: \[ \mathcal{L}(x, y, \lambda) = 5xy + \lambda (30 - 5x - y) \] 2. **Find the partial derivatives and set them to zero**: \[ \frac{\partial \mathcal{L}}{\partial x} = 5y - 5\lambda = 0 \quad \Rightarrow \quad y = \lambda \] \[ \frac{\partial \mathcal{L}}{\partial y} = 5x - \lambda = 0 \quad \Rightarrow \quad \lambda = 5x \] \[ \frac{\partial \mathcal{L}}{\partial \lambda} = 30 - 5x - y = 0 \quad \Rightarrow \quad y = 30 - 5x \] 3. **Substitute \( \lambda \) from the first equation into the second equation**: \[ y = 5x \] 4. **Substitute \( y = 5x \) into the budget constraint**: \[ 5x + 5x = 30 \] \[ 10x = 30 \] \[ x = 3 \] 5. **Find the corresponding \( y \)**: \[ y = 5x = 5 \times 3 = 15 \] 6. **Find the Lagrange multiplier \( \lambda \)**: From the equation \( \lambda = 5x \): \[ \lambda = 5 \times 3 = 15 \] So, the Lagrange multiplier associated with the equality constraint is \( \lambda = 15 \). The correct answer is: - \( 15 \)

To solve the utility maximization problem given the utility function \( u(x, y) = 5xy \), the budget constraint \( 5x + y = 30 \), and non-negative consumption of goods \( x \) and \( y \), follow these steps: 1. **Set up the Lagrangian**: \[ \mathcal{L}(x, y, \lambda) = 5xy + \lambda (30 - 5x - y) \] 2. **Find the partial derivatives and set them to zero**: \[ \frac{\partial \mathcal{L}}{\partial x} = 5y - 5\lambda = 0 \quad \Rightarrow \quad y = \lambda \] \[ \frac{\partial \mathcal{L}}{\partial y} = 5x - \lambda = 0 \quad \Rightarrow \quad \lambda = 5x \] \[ \frac{\partial \mathcal{L}}{\partial \lambda} = 30 - 5x - y = 0 \quad \Rightarrow \quad y = 30 - 5x \] 3. **Substitute \( \lambda \) from the first equation into the second equation**: \[ y = 5x \] 4. **Substitute \( y = 5x \) into the budget constraint**: \[ 5x + 5x = 30 \] \[ 10x = 30 \] \[ x = 3 \] 5. **Find the corresponding \( y \)**: \[ y = 5x = 5 \times 3 = 15 \] So, the x-coordinate of the point that solves this individual's utility maximization problem is \( x = 3 \). The correct answer is: - \( 3 \)

A consumer spends an amount 𝑚𝑚 to buy 𝑥𝑥 units of one good at the price of 6 per unit and 𝑥𝑥 units of a different good at the price of 10 per unit. Here 𝑚𝑚 is positive and suppose that 8 < 𝑚𝑚 < 40. The consumer’s utility function is 𝑈(𝑥, y) = 𝑥y + y^2+ 2𝑥 + 2y, so that her problem is: Maximize 𝑥y+y^2 +2x+2y subject to 6𝑥+10𝑥y=𝑚 (a) Find the optimal quantities 𝑥∗ and y∗ and the Lagrange multiplier, all of them as functions of 𝑚. (b) Write down the maximum value of the utility function as a function of m. (c) Find the derivative of the above utility function for m=20 (d) What are the solutions for 𝑥∗ and y∗ if (i) 𝑚 ≤ 8? (ii) 𝑚 ≥ 40?

Apply the Lagrange multiplier for a firm that faces the production function Q = 2K0.2 L0.6 and can buy L at R240 a unit and K at R4 a unit.(a) If it has a budget of R16,000 what combination of K and L should it use to maximize output? (7)(b) If it is given a target output of 40 units of Q what combination of K and L should it use to minimize the cost of this output?

Let T(x,y,z) = x^2 + y^2 + z^2 and h(x,y,z) = 2x + 3y - 5z + 4. Which one of the following systems of equations represents the Lagrange multiplier condition that must be satisfied by a point that maximises or minimises T subject to h(x,y,z) = 0?

Use the method of Lagrange Multipliers to optimise the function f (x, y) = x2 + y2 − 3xy subject to the constraint x − 4y = 7. Find the location of the optimal point, the value of λ and the value of f (x, y) at this point. Note: In the quiz, enter non-integer numerical values as decimals to at least 3 decimal places; i.e. you will need to fill in some boxes similar to the following: The optimal point is located at x = and y = with λ = and f (x, y) =