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Could anyone guide me on how to plot the function VR with the values of ρ that I get from solving the following eq.

eq = (2 - 7 a) + (1 - ρ)^(-1 - b) ρ (2 - 2 ρ + b ρ) == 0;

I'm trying to first plot the above function and based on the value of ρ I want to plot VR function which is:

10 (10 a*ρ + ρ (-2 - 3 a - 2 (1 - ρ^(-b) ρ))

for plotting the eq I'm doing the following

0<=a<=1 ; 0 <= ρ <= 1; 0 <= b <= 3
 eq = (2 - 7 a) + (1 - ρ)^(-1 - b) ρ (2 - 2 ρ + b ρ) == 0;
soln = First@ Solve[eq, a];
Show[ParametricPlot3D[{a /. soln, b, ρ}, {b, 0, 3}, {ρ, 0, 1}, 
   PlotRange -> {{0, 1}, {0, 3}, {0, 1}}, 
   AxesLabel -> (Style[#, 16] & /@ {"a", "b", "ρ"})], 
 ParametricPlot3D[{a, b, 0}, {a, 0, 2/7}, {b, 0, 3}]]

However, i dont know how to plot VR based on the values of ρ

Thanks

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Parentheses are unbalanced in the second expression. So, let us assume that it is meant to be

f = 10 a*ρ + ρ (-2 - 3 a - 2 (1 - ρ^(-b) ρ))

If not, it can be corrected and the procedure below (edited based on OP's comment below) followed:

ParametricPlot3D[{a /. soln, b, f /. soln}, {b, 0, 3}, {ρ, 0, 1},
    PlotRange -> {{0, 1}, {0, 3}, {0, 1}}, 
    AxesLabel -> (Style[#, 16] & /@ {"a", "b", "VR"})]

where soln is obtained from 83393.

enter image description here

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  • $\begingroup$ thanks but I want to plot VR instead of ρ (i need to have a, b and VR in my graph). So if I solve eq from soln = First@ Solve[eq, a]; then how can I plot VR (of f in this case) based on the values of ρ ? $\endgroup$ – Baboda May 16 '15 at 13:56
  • $\begingroup$ I modified the solution per your request a few days ago. Does it meet your needs? $\endgroup$ – bbgodfrey May 21 '15 at 1:29
  • $\begingroup$ thanks, but I'm not sure if this is the correct way. because the equation is for obtaining the optimal ρ, which then those optimal solution must be used to plot VR function. so im not sure solving the eq for a is the right way. $\endgroup$ – Baboda May 21 '15 at 12:20
  • $\begingroup$ Plug in some numbers and see for yourself whether it is correct. $\endgroup$ – bbgodfrey May 21 '15 at 12:22

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