I have tried another one. Again I 'm getting the notification that, 'Not enough input arguments'. Can you help me to solve.

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shunmugam hemalatha 2021-5-31

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https://ww2.mathworks.cn/matlabcentral/answers/844165-i-have-tried-another-one-again-i-m-getting-the-notification-that-not-enough-input-arguments-ca

评论： shunmugam hemalatha 2021-6-2

clc
clear all
a=120; 
b=100; 
k1=4; 
k2=80; 
h=0.33; 
theta=0.07; 
delta=0.5; 
c2=2.5; 
c3=10; 
c4=5;
syms T t
C = (1/T)*(((t^3)*(((a*h)/(6*theta))+((c2*a*theta)/2)))+((b/2)*(c3+(delta*c4))*((T-t)^2))+(k1*t)+k2);
%[T,t]=solve(C)
Vars = [T, t];
[C]= solve(var)
%S= solve (C, T, t)

To get the answer t= 15228, T = 1.5871 and C =80.2626.

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Torsten 2021-5-31

If you want a numerical answer, use

S = double(solve(CC==0,var))

shunmugam hemalatha 2021-5-31

I'm very sorry to answer . Still, I'm getting

S =

-8.4285 + 0.0000i

1.2320 - 0.5961i

1.2320 + 0.5961i

Actually, I need the procedure to calculate the two variables in one equation. Will you please to get the answer as

t= 15228, T = 1.5871 and C =80.2626.

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回答（1 个）

Walter Roberson 2021-5-31

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https://ww2.mathworks.cn/matlabcentral/answers/844165-i-have-tried-another-one-again-i-m-getting-the-notification-that-not-enough-input-arguments-ca#answer_713395

编辑：Walter Roberson 2021-5-31

在 MATLAB Online 中打开

Vars = [T, t];
[C]= solve(var)

You ask to solve(var) but var is not defined by your code, so var is being taken as a reference to the variance function var()

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Walter Roberson 2021-5-31

在 MATLAB Online 中打开

If you define

syms T t C
eqn = C == (1/T)*(((t^3)*(((a*h)/(6*theta))+((c2*a*theta)/2)))+((b/2)*(c3+(delta*c4))*((T-t)^2))+(k1*t)+k2)

then you would have a single equation in three variables. You would be unlikely to find a single solution -- any change in T or t would result in a different right hand side, and you could call C whatever the new result was, since C does not appear on the right hand side.

If you want to solve for T t C then you need three simultaneous equations, except cases where you can prove that the equations have no solutions except at some special points (for example, if there were a contradiction that only vanished if T were infinity so that 1/T was 0 causing the rest of the expression to vanish.)

shunmugam hemalatha 2021-6-2