Rhoda takes a job with a construction company.She earns $5 an hour for the first 40 hours of each week and then gets "double-time" for overtime.That is, she is paid $10 an hour for every hour beyond 40 hours a week that she works.Rhoda has 70 hours a week available to divide between construction work and leisure.She has no other source of income, and her utility function is U  cr, where c is her income to spend on goods and r is the number of hours of leisure that she has per week.She is allowed to work as many hours as she wants to.How many hours will she work?

A) 50
B) 30
C) 45
D) 35
E) None of the above.

Rhoda's goal is to maximize her utility, subject to the constraint that she has 70 hours per week to allocate between work and leisure. Let F1 be the number of hours she works and S1 be the number of hours she spends on leisure. Then, we have:
F1 + S1 = 70 (constraint)
Her income, I, is given by:
I = 5F1 + 10 max(0,F1-40) (wage function)
We can simplify this as follows:
I = 5F1 + 10(F1-40) (since max(0,F1-40) = F1-40 for F1 >= 40)
I = 15F1 - 400
Her utility function is:
U = F1 * S1^0.5
We want to choose F1 and S1 to maximize U subject to the constraint on time:
Maximize U = F1 * S1^0.5
Subject to F1 + S1 = 70
We can use Lagrange multipliers to solve this optimization problem:
L(F1,S1,lambda) = F1 * S1^0.5 + lambda(70 - F1 - S1)
Taking partial derivatives with respect to F1, S1, and lambda and setting them equal to zero, we get:
S1^0.5 = lambda
0.5 * F1 * S1^-0.5 - lambda = 0
70 - F1 - S1 = 0
From the first equation, we have S1 = lambda^2. Squaring the third equation gives:
F1^2 - 140F1 + 4900 = 0
Solving for F1 using the quadratic formula, we get:
F1 = 70 - S1 = 45 (rounded to the nearest integer)
So Rhoda should work 45 hours and spend 25 hours on leisure to maximize her utility.