How to maximise the objective function with constraints and plot the solution

I want to maximise the objective function with respect to x

function=a*ln(1+x)+b(c-x)

subjected to constraints : x,min<=x<=x,max

given parameters are a,b,c.(Randamly I choose some values but it shows error as 'Index exceeds the number of array elements. Index must not exceed 1",how to correct this?)

Also plot the x verses function graph.

Answer 1

[William Rose]

@Ancy S G, Since this is a simple function of one adjustable variable, use calculus: differentiate and solve for where the derivative is zero.

therefore

which equals zero when

.

Test if . If x_opt is NOT in that range, then the value that maximizes the function inside the range will be at one end or the other of the range, so simply evaluate both endpoints and pick the higher one.

To demonstrate that this is true, let's pick random values for a,b,c. Then we will compute x_opt and y(x_opt). Then we will plot y(x) over a range to see if we got the maximum right.

Try it.

[William Rose]

Code for above idea

a=.3; b=.65; c=.1;  %random values for a,b,c
xopt=a/b-1;         %theoretical location of the maximum
yopt=a*log(1+xopt)+b*(c-xopt);  %theoretical max value
x=-.95:.05:10;          %x values for plotting
y=a*log(1+x)+b*(c-x);   %y values for plotting
plot(xopt,yopt,'rx',x,y,'-r');  %plot max value with an X and a line for all points

fprintf('a=%.2f,b=%.3f,c=%.3f,xopt=%.3f,yopt=%.3f\n',a,b,c,xopt,yopt);

a=0.30,b=0.650,c=0.100,xopt=-0.538,yopt=0.183

Try.

[Torsten]

Note that x_opt = a/b -1 will be a local minimum if a < 0.

So in your analysis, you'll also have to check whether a is positive or negative.

[Ancy S G]

Sir,I have a doubt.Can we find the optimal point of this function on the graph without providing the theoretical maximum value?

[William Rose]

@Ancy S G, Yes. Use fminbnd() to minimize a function of one variable. See attached code.

In the script, at the bottom, I define function ancy1(x,a,b,c). In the main body of the code, I use fminbnd() to find the minimum of -ancy1() with respect to x. Then I take the negative of that, to get the maximum of +ancy1(). Results are displayed graphically and on the console. The maximum found by calculus matches the minimum found numerically.

Answer 2

[William Rose]

@Ancy S G,

The routine fmincon() is good for this. The name is short for function minimization with constraints. Minimize the negative of your function in order to maximize.

[Ancy S G]

Ok.Thank you very much

[William Rose]

@Ancy S G, you're welcome.