sol16-10

# sol16-10 - SOLUTIONS FOR HOMEWORK PROBLEMS 16.10 MATH 132 -...

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Unformatted text preview: SOLUTIONS FOR HOMEWORK PROBLEMS 16.10 MATH 132 - WI00 4 Equilibrium: 400- q 2 = 20 q + 100 → 0 = q 2 + 20 q- 300 → ( q- 10)( q + 30) = → q =- 30 and q = 10. We drop the negative value for practical reasons, so we’re stuck with q = 10 → p = 20 * 10 + 100 = 300. The surpluses now will basically just be each of these functions, out of which we substract the equilibrium p (100) integrated from 0 to equilibrium q ; if you compute the integrals like that, one integral will come out positive, and one negative - the positive one is the Consumer’s Surplus, since it’s the one above all, and the negative one is, without the negative sign, the Producer’s Surplus, since the producer’s graph is theone below all; but one trick: the Consumer’s function will almost always have negative highest term (in our case- q 2 ) - exception will make case in which q appears in the denominator, for example - as for the Producer, it will almost always have positive highest term (20 q ), so you can see right away (the problem might even give you this information,...
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## This note was uploaded on 07/26/2011 for the course MATH 132 taught by Professor Staff during the Spring '08 term at Ohio State.

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sol16-10 - SOLUTIONS FOR HOMEWORK PROBLEMS 16.10 MATH 132 -...

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