An open channel has a trapezoidal cross-section as shown in the figure. If all the sides have an equal length of 2 m, find the angle x that maximizes the cross-sectional area using: a) A Graphical Approach b) The Golden Section Search (with es=0.05)

"An open channel has a trapezoidal cross-section as shown in the figure. If all the sides have an equal length of 2 m, find the angle x that maximizes the cross-sectional area using: a) A Graphical Approach b) The Golden Section Search (with es=0.05)"

I did this :

xl=0;
xu=180;
ea=10;
es=0.05;
f=@(x) 4.*sind(x)+4.*cosd(x).*sind(x);
d=(1.61803-1)*(xu-xl);
x1=xl+d;
x2=xu-d;

for i=0;  
  f(x2); f(x1);
  if f(x2) > f(x1)
    xopt= x1;
    x1 = x2;
    d=(1.61803-1)*(xopt-xl);
    x2=xopt-d;
  end
  
  if f(x1)< f(x2)
    xopt = x2;
    x2=x1;
    d=(1.61803-1)*(xu-xopt);
    x1=xopt+d;
  end
  if ea>es    
    ea = (2 - 1.61803) * abs((xu - xl) / xopt)*100 ;
  end  
end
i=1+i
xopt

What is the problem with my code?

[Alexandra Harkai]

Attach the figure. Show what you have tried and where the problem is.

how-do-i-get-help-on-homework-questions-on-matlab-answers

Answer 1

[Image Analyst]

You know the height is y = 2 * sin(theta).

The rectangular center's area is 2 * y.

The Triangle's base is x = 2 * cos(theta);

The area of each triangular side is (1/2)*x*y, and there are two of them,

so the total area = 2*y+x*y

Substitute x and y for the formula in terms of theta. Then make theta equal

theta = linspace(0, pi/2, 1000);

Then plot a graph of area as a function of theta. You can get the max area with the max() function. Be sure to use both output arguments of max() so you can use the index to get the theta where the max occurs.