An economics student faces the following utility function:

U=100W+150L+0.78WL−2.5W2−0.708L2

where $W$ is hours worked per week in order to raise income for course fees (as well as chips and ale) and $L$ is all other non-paid work time (leisure and including studying time). How much W
 and L
 must the student undertake in order to maximise her utility?   

To find the values of $ W $ (hours worked) and $ L $ (non-paid work time) that maximize the student’s utility, we need to:

1. Take partial derivatives of the utility function $ U $ with respect to $ W $ and $ L $.
2. Set these partial derivatives to zero to find critical points.
3. Solve the resulting system of equations.
4. Verify the solution represents a maximum using the second-order conditions (check the Hessian matrix).

The utility function is:
\[
U = 100W + 150L + 0.78WL - 2.5W^2 - 0.708L^2
\]

### Step 1: Compute Partial Derivatives
#### Partial derivative with respect to $ W $:
\[
\frac{\partial U}{\partial W} = 100 + 0.78L - 5W
\]

#### Partial derivative with respect to $ L $:
\[
\frac{\partial U}{\partial L} = 150 + 0.78W - 1.416L
\]

### Step 2: Set Derivatives to Zero
Set $\frac{\partial U}{\partial W} = 0$ and $\frac{\partial U}{\partial L} = 0$:
\[
100 + 0.78L - 5W = 0 \quad \text{(1)}
\]
\[
150 + 0.78W - 1.416L = 0 \quad \text{(2)}
\]

### Step 3: Solve the System of Equations
From equation (1):
\[
0.78L = 5W - 100
\]
\[
L = \frac{5W - 100}{0.78} \quad \text{(3)}
\]

Substitute equation (3) into equation (2):
\[
150 + 0.78W - 1.416 \left(\frac{5W - 100}{0.78}\right) = 0
\]
Simplify the terms:
\[
150 + 0.78W - \frac{1.416(5W - 100)}{0.78} = 0
\]
\[
150 + 0.78W - \frac{7.08W - 141.6}{0.78} = 0
\]
\[
150 + 0.78W - (7.08W - 141.6) = 0
\]
\[
150 + 0.78W - 7.08W + 141.6 = 0
\]
\[
291.6 - 6.3W = 0
\]
\[
W = \frac{291.6}{6.3} \approx 46.29
\]

Substitute $ W = 46.29 $ into equation (3) to find $ L $:
\[
L = \frac{5(46.29) - 100}{0.78}
\]
\[
L = \frac{231.45 - 100}{0.78} \approx \frac{131.45}{0.78} \approx 168.27
\]

### Step 4: Verify Second-Order Conditions
The second-order partial derivatives are:
\[
\frac{\partial^2 U}{\partial W^2} = -5, \quad \frac{\partial^2 U}{\partial L^2} = -1.416, \quad \frac{\partial^2 U}{\partial W \partial L} = 0.78
\]

The Hessian matrix is:
\[
H = \begin{bmatrix}
-5 & 0.78 \\
0.78 & -1.416
\end{bmatrix}
\]

The determinant of $ H $ is:
\[
\text{Det}(H) = (-5)(-1.416) - (0.78)^2 = 7.08 - 0.6084 = 6.4716
\]

Since $\text{Det}(H) > 0$ and $\frac{\partial^2 U}{\partial W^2} < 0$, the utility function has a maximum at $ W \approx 46.29 $ and $ L \approx 168.27 $.

### Final Answer:
To maximize utility:
\[
W \approx 46.29 \, \text{hours worked per week, and} \, L \approx 168.27 \, \text{hours of leisure/non-paid work}.
\]

Would you like me to provide further details or clarify any step? Here are some related questions:

1. How does the Hessian matrix confirm whether a critical point is a maximum or minimum?  
2. What are the economic interpretations of $ W $ and $ L $ in this utility function?  
3. How does changing coefficients in the utility function affect the results?  
4. What happens if the second-order conditions are not satisfied?  
5. Can constraints (e.g., total hours available) be incorporated into this problem?

**Tip:** When solving optimization problems, always

An economics student faces the following utility function: U = 100W + 150L + 0.78WL − 2.5W^2 − 0.708L^2 where W is hours worked per week in order to raise income for course fees (as well as chips and ale) and L is all other non-paid work time (leisure and including studying time). How much W and L must the student undertake in order to maximize her utility?

Learn how to maximize utility using partial derivatives and the Hessian matrix. This problem explores the trade-off between work and leisure time with the utility function U = 100W + 150L + 0.78WL − 2.5W^2 − 0.708L^2. Includes detailed step-by-step calculations for finding W and L that maximize utility.

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