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{SECT 0 {PARA 0 "" 0 "" {TEXT 23 8 "Econ 407" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 23 83 "Assignment: Inst. N. hash: Appl
ications of derivatives: how to find the minimum and" }}{PARA 0 "" 0 "
" {TEXT 23 44 "maximum of  cost and revenue functions." }}{PARA 
0 "" 0 "" {TEXT 23 0 "" }}{PARA 0 "" 0 "" {TEXT 23 75 "1.The cost of p
roducing q items is given by C = 15q^(2/3) + 100q + 10,000. " }}{PARA 
0 "" 0 "" {TEXT 267 53 "Find where this cost function is increasing an
d where" }}{PARA 0 "" 0 "" {TEXT 23 39 "it is decreasing. Sketch the g
raph of C" }}{PARA 0 "" 0 "" {TEXT 23 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "C:=15*q^(2/3) + 100*q + 10000;" }}{PARA 6 "" 1 "" 
{TEXT -1 27 "C := 15*q^(2/3)+100*q+10000" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "plot(C,q = 0..40000);" }}{PARA 13 "" 1 "" {GLPLOT2D 
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284 "As the plot of the cost curve shows, it is an increasing function
 of output--q. We can also take the first derivative and evaluate the \+
first derivative for various values of q. But the rate of increase in \+
the cost function decreases. In other words, it increases at a decreas
ing rate." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 14 "dc:=diff(C,q);" }}{PARA 6 "" 1 "" {TEXT -1 20 "dc :
= 10/q^(1/3)+100" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eval(dc
, q=100.0);" }}{PARA 6 "" 1 "" {TEXT -1 11 "102.1544347" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "eval (dc, q= 1000.0);" }}{PARA 6 "
" 1 "" {TEXT -1 11 "101.0000000" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 22 "plot(dc, q=0...43000);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 23 76 " I
f a company sells q items per day, then the price of each item is give
n by" }}{PARA 0 "" 0 "" {TEXT 256 0 "" }}{PARA 0 "" 0 "" {TEXT 23 13 "
P = 10 - q/10" }}{PARA 0 "" 0 "" {TEXT 23 0 "" }}{PARA 0 "" 0 "" 
{TEXT 23 80 "Find the range where the revenue function R(q) =qp is inc
reasing and where it is" }}{PARA 0 "" 0 "" {TEXT 23 11 "decreasing." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 18 "TR:=(10.0-q/10)*q;" }}{PARA 6 "" 1 "" {TEXT -1 21 "TR := (10.0-1
/10*q)*q" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(TR, q=0..1
20);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVE
SG6$7S7$\"\"!F(7$$\"1++++&ech#!#:$\"1xP61=CZD!#97$$\"1+++v*G:*[F,$\"1a
t'yTeAl%F/7$$\"1+++]L)4X(F,$\"1F-=@=\"e*oF/7$$\"1+++X.u-5F/$\"1#G40D:>
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FD7$$\"1++]FPQ^AF/$\"1?[le3^W<FD7$$\"1+++IrS7DF/$\"1;DJr\")=\")=FD7$$
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%**)F/$\"1Bx%ROzm/*F/7$$\"1,+]Z=iY#*F/$\"1&oNV)G?mpF/7$$\"1******\\'[M
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0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 251 "As can be seen from the graph, TR increases  until outpu
t reaches about 50 units, then it begins to drop off. Beyond q=100, TR
 becomes negative.  We can find the maximum point on TR by taking its \+
derivative and setting equal to zero and solving for q." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 15 "MR:=diff(TR,q);" }}{PARA 6 "" 1 "" {TEXT -1 
17 "MR := -1/5*q+10.0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "so
lve(MR=0,q);" }}{PARA 6 "" 1 "" {TEXT -1 3 "50." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 23 0 "" }}{PARA 0 "
" 0 "" {TEXT 23 75 "3. The cost of producing q items is given by C(q) \+
= 100 +50q + 0.2q^3. Find" }}{PARA 0 "" 0 "" {TEXT 23 75 "dC/dq and d^
2C/dq^2 at q = 0, 5, 50, and 75. Find the first and the second " }}
{PARA 0 "" 0 "" {TEXT 257 78 "derivative of the cost function. The sec
ond derivative is a useful concept and" }}{PARA 0 "" 0 "" {TEXT -1 43 
"will be discussed in class at a later date." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 23 0 "" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "C:= 100 +50*q + 0.2*q^3;" }}{PARA 6 "" 1 "" {TEXT -1 
20 "C := 100+50*q+.2*q^3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 
"dc:=diff(C,q);" }}{PARA 6 "" 1 "" {TEXT -1 15 "dc := 50+.6*q^2" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "d2c:=diff(dc,q);" }}{PARA 6 
"" 1 "" {TEXT -1 12 "d2c := 1.2*q" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "plot(C, q=150..300);" }}{PARA 13 "" 1 "" {GLPLOT2D 
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}}{PARA 0 "" 0 "" {TEXT 23 0 "" }}{PARA 0 "" 0 "" {TEXT 23 67 "4. The \+
postage for shipping an item weighing q ounces is modeled by" }}{PARA 
0 "" 0 "" {TEXT 258 0 "" }}{PARA 0 "" 0 "" {TEXT 23 0 "" }}{PARA 0 "" 
0 "" {TEXT 23 19 "P = Q(Q^2 +1)^(0.5)" }}{PARA 0 "" 0 "" {TEXT 23 0 "
" }}{PARA 0 "" 0 "" {TEXT 23 50 "Find the dP/dq and d^2P/dq2 when q = \+
1, 10 and 20." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 22 "P:=(q)*(q^2 +1)^(0.5);" }}{PARA 6 "" 1 "" {TEXT 
-1 17 "P := q*(q^2+1)^.5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 
"plot(P,q=200..1000);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
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}}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "dp:= diff(P,q);" }}
{PARA 6 "" 1 "" {TEXT -1 35 "dp := (q^2+1)^.5+1.0*q^2/(q^2+1)^.5" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "ddp:=diff(dp,q);" }}{PARA 6 
"" 1 "" {TEXT -1 44 "ddp := 3.0/(q^2+1)^.5*q-1.00*q^3/(q^2+1)^1.5" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eval(dp,q=1);" }}{PARA 6 "" 
1 "" {TEXT -1 11 "2.121320343" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 14 "eval(ddp,q=1);" }}{PARA 6 "" 1 "" {TEXT -1 11 "1.767766953" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "eval(dp=10);" }}{PARA 6 "" 
1 "" {TEXT -1 34 "(q^2+1)^.5+1.0*q^2/(q^2+1)^.5 = 10" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 15 "eval(ddp,q=10);" }}{PARA 6 "" 1 "" {TEXT 
-1 11 "1.999926234" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 23 0 "" }}{PARA 
0 "" 0 "" {TEXT 23 78 "5. Suppose the daily profit (in dollars) from m
anufacturing q items is modeled" }}{PARA 0 "" 0 "" {TEXT 260 75 "by P(
q)= -0.01q^2 + 5q-400, where 0<q<300. Find the number of items q that \+
" }}{PARA 0 "" 0 "" {TEXT 261 27 "results in maximum profits." }}
{PARA 0 "" 0 "" {TEXT 23 0 "" }}{PARA 0 "" 0 "" {TEXT 23 75 "Take the \+
derivative of the profit function and set equal to zero and solve " }}
{PARA 0 "" 0 "" {TEXT -1 6 "for q." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 23 72 "6. Suppose the cost (in dollars) of manuf
acturing q items is modeled by " }}{PARA 0 "" 0 "" {TEXT 262 73 "C =40
0+ 10q +0.01q^2. Find the value of q that minimizes the average cost" 
}}{PARA 0 "" 0 "" {TEXT 263 75 "function. Solution: first compute AC f
unction. This is done by dividing the" }}{PARA 0 "" 0 "" {TEXT 264 49 
"total cost function by output--q, as shown below." }}{PARA 0 "" 0 "" 
{TEXT 265 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "C:=400.0+ \+
10*q +0.01*q^2;" }}{PARA 6 "" 1 "" {TEXT -1 25 "C := 400.0+10*q+.1e-1*
q^2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "AC:=C/q;" }}{PARA 6 "
" 1 "" {TEXT -1 30 "AC := (400.0+10*q+.1e-1*q^2)/q" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 20 "plot(AC,q=100..350);" }}{PARA 13 "" 1 "" 
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 172 "Looking at the graph, the functio
n shows a minimum around q=200. To be more exact, we take the first de
rivative of the AC function with respect to q, and set equal to zero.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "dAC:=diff(AC,q);" }}{PARA 6 "" 1 "" {TEXT -1 48 "dAC \+
:= (10+.2e-1*q)/q-(400.0+10*q+.1e-1*q^2)/q^2" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 13 "solve(dAC,q);" }}{PARA 6 "" 1 "" {TEXT -1 12 "20
0.   -200." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "So the solution is \+
at q=200." }}}{PARA 0 "" 0 "" {TEXT 23 0 "" }}}{MARK "0 0" 8 }
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