Re: graphs and AxesOrigin
- To: mathgroup at smc.vnet.net
- Subject: [mg43095] Re: graphs and AxesOrigin
- From: bobhanlon at aol.com (Bob Hanlon)
- Date: Tue, 12 Aug 2003 04:43:08 -0400 (EDT)
- References: <bh7d77$54n$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
c=0.5;
M=1000;
k1=950;
d=0.52;
k2=750;
p2=(1-c)*((M-d*(M-k2))/(2*(1-d)*k1))+c;
S1=(1-c)*((1-d)*k1+d*(M-k2));
L1=(1-c)*(((1-d)*(p2-c)*k1+d*(1-c)*(M-k2))/(1-c));
U1=(1-c)*(d*k1+(1-d)*(M-k2));
f[v_]:=Min[(1-c)*M-v,
((v-d*(1-c)*(M-k2))/
((1-d)*k1))*(M-v/(1-c))];
Plot[f[v],{v,L1,U1},
PlotStyle->RGBColor[0,0,1],
AxesOrigin->{L1,f[U1]}];
You might want to check your definition for L1. As is, you appear to multiply
and divide by (1-c).
Bob Hanlon
In article <bh7d77$54n$1 at smc.vnet.net>, Emmanuel Dechenaux
<dechenau at purdue.edu> wrote:
<< I have what I assume is a simple question. I'm trying to make the graph
below but if I let mathemtica determine the origin it puts the vertical
axis in the middle of the curve. I don't want this to happen, so I use
AxesOrigin. Now unfortunately it doesn't show the axes all the way to
the origin, it doesn't go beyong the range I specified for the variable
in the function I want to plot. I was wondering if there's a way for
this not to happen (the plot range can't possibly be different). Thanks.
Emmanuel
\!\(c = 0.5\[IndentingNewLine]
M = 1000\[IndentingNewLine]
k1 = 950\[IndentingNewLine]
d = 0.52\[IndentingNewLine]
k2 = 750\[IndentingNewLine]
p2 = \((1 - c)\) \(M - d \((M - k2)\)\)\/\(2 \((1 - d)\) k1\) +
c\[IndentingNewLine]
S1 = \((1 - c)\) \((\((1 - d)\) k1 + d \((M - k2)\))\)\[IndentingNewLine]
L1 = \((1 -
c)\) \((\(\((1 - d)\) \((p2 - c)\) k1 + d \((1 - c)\) \((M - \
k2)\)\)\/\(1 - c\))\)\[IndentingNewLine]
U1 = \((1 -
c)\) \((d \((k1)\) + \((1 - d)\) \((M -
k2)\))\)\[IndentingNewLine]
Plot[Min[\((1 - c)\) M -
v, \((\(v - d \((1 - c)\) \((M - k2)\)\)\/\(\((1 - d)\) k1\))\)
\((M \
- v\/\((1 - c)\))\)], {v, L1, U1}]\)