Solving a student’s problem

Suppose that a student’s preference on grade and leisure can be represented by a continuous and concave utility function, U(grade, leisure) where it is an increasing function in each argument. Suppose further that student grade is an increasing function of how much time spent on reading i.e. grade = f(reading). The more the student reads, the higher grade he will obtain. The student must allocate their time in one day (24 hours) for reading and leisure. Hence the student faces the time constraint as reading + leisure ≤ 24. From this information, we can formulate the student problem as follow. 

Max            U(grade, leisure)

Subject to   reading + leisure ≤ 24

and             grade = f(reading) 

It can be shown that the time constraint is a close and bounded set. Since the utility is continuous function, from Weisrstrass’s Theorem, we know that there exists a solution to this problem. (See Mathematics for Economists by Simon&Blume (1994) page 823 for the proof of Weisrstrass’s Theorem.) 

However, we do not assume that the utility function satisfies the condition that ∂U(0, leisure)/∂grade →∞. Hence, we possibly end up with a boundary solution. Consider the following first order conditions. 

(∂U(.)/∂grade)x(∂grade/∂reading) – λ ≤ 0, and reading = 0

∂U(.)/∂leisure – λ = 0, and leisure > 0

reading + leisure = 24 

where  λ is a Lagrange multiplier for the time constraint. These are3 equations with 3 unknown (reading, leisure, λ) from which we can solve for the optimal decision on reading and leisure. Note that the first two of first order conditions imply 

(∂U(.)/∂grade)x(∂grade/∂reading) ≤ ∂U(.)/∂leisure evaluated at reading = 0, and leisure = 24, or
marginal benefit of reading ≤ marginal benefit of leisure evaluated at reading = 0, and leisure =24 

This implies that this student will not spend his time on reading at all. He will spend his 24 hours a day solely on leisure!!!?

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