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Unformatted text preview: will choose the first option and convert everything yearly. Let S(t) be the amount of the load remaining at year t, then dS/dt = rS – k, where r=0.09 and k=800x12 = 9600 is the maximal annual payment. The equation can be rewritten as (S exp(rt))' =  k exp(rt). Integrating, S(t) exp(rt)  S(0) = (k/r) [ exp (rt)  1] We want S(20)=0 when S(0) is the maximal amount this buyer can afford to borrow. Thus S(0) = (k/r)[1exp(20r)] = $89,034.79. The total interest paid is k x 20 – S(0) = $102,965.21 2.5 (page 88): 15...
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This note was uploaded on 01/29/2011 for the course MATH 215 taught by Professor Keqinliu during the Fall '08 term at UBC.
 Fall '08
 KEQINLIU
 Slope

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