Question

Check my we Use the golden-section method to solve for the value of x that maximizes 14--1.5X6-2/4 + 12x Employ initial guesses of xy0 and Xu-2, and perform three iterations. (Round the final answer to four decimal places.) The value of x that maximizes the given function is

Answer 1

SOLUTION :

Matlab code:

a=0;                            % start of interval
b=2;                            % end of interval
epsilon=0.000001;               % accuracy value
iter= 3;                       % maximum number of iterations
tau=double((sqrt(5)-1)/2);      % golden proportion coefficient, around 0.618
k=0;                            % number of iterations
function r = f(x)
r = -1.5*x^6 - 2*x^4 + 12*x;
end

x1=a+(1-tau)*(b-a);             % computing x values
x2=a+tau*(b-a);

f_x1=f(x1);                     % computing values in x points
f_x2=f(x2);

while ((abs(b-a)>epsilon) && (k<iter))
    k=k+1;
    if(f_x1>f_x2)
        b=x2;
        x2=x1;
        x1=a+(1-tau)*(b-a);
    
        f_x1=f(x1);
        f_x2=f(x2);
    else
        a=x1;
        x1=x2;
        x2=a+tau*(b-a);
    
        f_x1=f(x1);
        f_x2=f(x2);
    end

    sprintf('Iteration %d: x1=%f, f(x1)=%f, x2=%f, f(x2)=%f', k, x1, f_x1, x2, f_x2)
end

% chooses maximum point
if(f_x1<f_x2)
    sprintf('x_max=%f', x2)
    sprintf('f(x_max)=%f ', f_x2)
else
    sprintf('x_max=%f', x1)
    sprintf('f(x_max)=%f ', f_x1)
end

Output:

Iteration 1: x1=0.472136, f(x1)=5.549637, x2=0.763932, f(x2)=8.187885
Iteration 2: x1=0.763932, f(x1)=8.187885, x2=0.944272, f(x2)=8.677842
Iteration 3: x1=0.944272, f(x1)=8.677842, x2=1.055728, f(x2)=8.107398
x_max=0.944272
f(x_max)=8.677842

Hence max value of f(x) rounded upto 4 decimals is 8.6778