6.2#1
a) Price - about 28; quantity - about 130. If you estimated a little
differently don't sweat it.
b) Consumer - about 14 squares, each worth $25 * 10, so $3500.
Producer - about 8 squares, so $2000.
c) $5500 all together.
#2
a) About 9 squares for the consumer, so $2250, and 10.5 squares for the
producer, so $2625. Together, $4875.
b) With price controls, consumer surplus goes down, producer goes up,
and total goes down.
#4
a) Setting f,g = $50, we can solve for q, and get consumers willing to buy about 87 at that price, whereas producers are ready to make 100. So there's oversupply and prices will probably go down.
b) We set f=g and try to solve for q. Algebraically it looks pretty hopeless, so after playing with it for hopefully not too long a while we look on a calculator to see where the graphs cross. This happens at about price = $48, quantity = 91, which fits with a).
c) The consumer integral is int(f dq) with q from 0 to 91, minus 48*91.
We know how to do this integral and it comes out to about 2096.
The producer surplus is 48*91 minus int(g dq) from 0 to 91.
Splitting into pieces, pulling out the constant 4, and using
the int(x^n) rule at n=1/2, we get about 1143. The interpretation
of the surpluses is what it always is - how much they saved/gained by
trading at the market price instead of their base price.
6.3#2
Presently: int(3000 e^-.06t), t from 0 to 15. It comes out to about $29,672. Future (the end of the 15 years): multiply that by e^(.06*15), and you get about $72,980.
#3
int((100+10t) e^(-0.05t)), t from 0 to 10. First, distribute the product over the sum. Then break the integral in two (and pull out constants). One of them is just an exponential, but the other is the tricky one I gave in class and on the web. Doing them individually, and adding, we get about $1148.
#4
a) i) int(5000 e^-.03t) from 0 to 4 is about $18,847.
ii) int(5000 e^-.1t) from 0 to 4 is about $16,484.
Note that it's lower - we're less impressed with being offered money in
the future, if we know we can just make it up from interest.
b) Multiplying these by e^(.03*4) and e^(.1*4), respectively, we get about $21,249 (at 3%) and $24,591 (at 10%). Note that the second one is now higher - if we get a better interest rate, then the money dumped in before a certain time builds to to more by that time.
#6
If we dump in money at rate R for 5 years, its future value is int(R e^(.06*(5-t)) ) from 0 to 5, which is R * (e^.3 - 1)/.06, or approximately R*5.83. (The way to remember this future-value integral: the money dumped in at time t gets to grow for 5-t years.) We want that to be $20K, so R*5.83 = $20,000, hence R is about $3430.
#8
The income stream _itself_ is s(t) = 50 e^-t (i.e. not constant sales, but exponentially dropping). So the present value is int( s * e^-.06t) from 0 to 2, which is int(50 e^(-1.06 t)) from 0 to 2. This we know how to do and get about $41,508.
Don't outsmart yourself on this one - just because s itself has an exponential in it, doesn't mean that you leave the other one (discounting the value of future payments) out of the integral or something.
#16
a) q(t) is the rate of extraction, i.e. how loud a big sucking sound we hear at time t. If it were constant - say, 10 million bbl./year - then we would all 100 million bbl. out in 10 years. We can ask how much has come out by time T, and then set that equal to 100 and solve for T. We did the details of this particular problem in class; the time T is about 10.6. It's not surprising that it's so close to 10 because the rate goes down so slowly.
b) If we make $10 a barrel (net profit), then this is asking for us to integrate 10*q*e^(-0.1t) between now and the T found in part (a). Said integral involves the x*e^x integral we did a few times in class and on the web page. The answer is quite gross algebraically, but plugging in the numbers it comes out to about $625 million.