Let's work through each part of this problem in detail:

Given Information:

• Utility function: $U = u(x, y) = x^3 y$
• Income (M): $600
• Price of $x$ ($P_x$): $10
• Price of $y$ ($P_y$): $1

Let's solve each part.

Part (a): Find Devin's Demand Functions for $x$ and $y$

To maximize utility given the budget constraint, we'll use the Lagrange multiplier method.

1. Objective function: $U(x, y) = x^3 y$

2. Budget constraint: $10x + 1y = 600$

3. Lagrangian function: $\mathcal{L} = x^3 y + \lambda (600 - 10x - y)$

4. Set up partial derivatives and solve for the demand functions $x$ and $y$.

   $\frac{\partial \mathcal{L}}{\partial x} = 3x^2 y - 10\lambda = 0$ $\frac{\partial \mathcal{L}}{\partial y} = x^3 - \lambda = 0$ $\frac{\partial \mathcal{L}}{\partial \lambda} = 600 - 10x - y = 0$

From the second equation, $\lambda = x^3$. Substitute this into the first equation:

$3x^2 y = 10x^3$

Dividing both sides by $x^2$ (assuming $x \neq 0$):

$3y = 10x$ $y = \frac{10}{3} x$

Now, substitute $y = \frac{10}{3} x$ into the budget constraint:

$10x + \frac{10}{3} x = 600$

Combine terms:

$\frac{30x + 10x}{3} = 600$ $\frac{40x}{3} = 600$ $40x = 1800$ $x = 45$

Now, substitute $x = 45$ back into $y = \frac{10}{3} x$:

$y = \frac{10}{3} \times 45 = 150$

So, the demand functions (values) are:

• $x = 45$
• $y = 150$

Part (b): Devin's Optimal Consumption Bundle and Total Utility

Devin's optimal consumption bundle, based on the demand functions we found, is:

• $x = 45$
• $y = 150$

Total Utility:

$U = x^3 y = 45^3 \times 150$

Calculate $45^3$:

$45^3 = 91125$

Now, multiply by 150:

$U = 91125 \times 150 = 13668750$

So, Devin's total utility at this bundle is 13,668,750.

Part (c): New Consumption Bundle and Total Utility if Price of $x$ Decreases to $5

1. Updated price of $x$: $P_x = 5$

2. New budget constraint: $5x + y = 600$

Using the same method as above, set up the new system to find the new demand functions:

With the utility maximization process similar to Part (a):

• The relationship $y = \frac{10}{3} x$ still holds.

Substitute $y = \frac{10}{3} x$ into the new budget constraint:

$5x + \frac{10}{3} x = 600$

Combine terms:

$\frac{15x + 10x}{3} = 600$ [ \frac{25x