Hello,

I have tried to evaluate my function for it's maximum, but I keep failing. What am I missing here?

I realize this is a very basic calculus problem, but my function I am determing the max output power for has it's own set of variables that I am optimizing, and I think that is why I cannot use the function "handle"? I am sure my terminology is incorrect.

I get the correct plot, but I just want to spit out the value of the maximum of the plot (ymax value), which in the code I have labeled as P_ave_output(R_l_optimal). Shouldn't this give me the max value of the function?

`clear allsyms omega t R_l s A_in Y%Inputs`

A_o = 3;R_c = 40;L_c = 0.051;f=186;omega_in = 2*pi*f;A = A_o*sin(omega*t);M=0.01;%Theta from part a)

theta = 0.244;%Spring coefficient (N/m)

k = 13660;%Damping coefficient (N*s/m)

C_d = 0.07;%Capacitance Assumption

C=12*10^-6;%Impedance equations

Z1 = R_c+R_l+L_c*s+1/(s*C);Z2 = C_d+k/s+M*s;%From part a) transfer function

%Output voltage V_l

V_l = (theta*(Y*s)*R_l)/Z1;%Input excitation force Z_in

A_in=(Y*s*Z2+theta*V_l/R_l)/(M);%output voltage ratio to input force

Transfer_func = simplify(V_l/A_in);Transfer_func(s) = vpa(simplify(V_l/A_in),5);Transfer_func(omega) = subs(Transfer_func, {s},{1j*omega});amplitude_TF = A_o*abs(Transfer_func(omega_in));angle_TF = angle(Transfer_func(omega_in));V_l_steady_state = amplitude_TF*(sin(omega_in*t+angle_TF));P_ave_output = (amplitude_TF/(sqrt(2)))^2/R_l;P_ave_diff = diff(P_ave_output);P_ave_diff = simplify(P_ave_diff, 'Steps',500); P_ave_max = P_ave_diff == 0;R_l_optimal = vpa(abs((solve(P_ave_max, R_l))),3)fplot(P_ave_output,[10 200])xticks([10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200])%xline(72.3)

ylabel('Average Power (Watts)')xlabel('Load Resistance R_l (Ohms)')P_ave_output(R_l_optimal)

## Best Answer