# We reduce our problem to maximizing 3 y p y over the

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, we reduce our problem to maximizing (3 - y ) /p + y over the set y 0. Note that the maximum must be at least 3 /p , which is the value taken at y = 0. It is easy to see that lim y →∞ (3 - y ) /p + y = -∞ , so there is some b with (3 - y ) /p + y < 3 /p for y > b . We can now remove any y > b from the set we are maximizing over, and focus on whether (3 - y ) /p + y can be maximized over [0 , b ]. But now we are maximizing a continuous function over a compact set, and Weierstrass’s Theorem tells us that there is a maximum. This is also a maximum for the original problem.
• Fall '08
• STAFF
• Economics, Topology, Metric space, λ, Closed set, mathematical economics midterm

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