derivative free optimaization method Currently I am working project on the derivative of free optimization methods. however, I want find practical problem  that solved using this method. So, how can I get solve practical examples using derivative free method? Please if you can, help me soon.
Thanks in advance.

Best regards,
Yohannes
I must confess, this is not derivative free, because derivatives make life easier.

From what I gather, you want to optimize an area given certain factors.  Let’s assume you were given a property that bordered a building, you would like to get the most sqft given x amount of fence.

First thing we need to do is turn this into a math problem.  We know two things about squares.  That the area A=LW, and the perimeter = 2L+2W.  given we only have 3 sides to our area in question, we assume that P= 2L+W. let’s assign X and Y instead.  Say we have 100 feet of fencing

P=2X+Y; 100=2X+Y.  let’s solve for Y
100=2X+Y
100-2X=2X+Y-2X
Y=100-2x

Now plug that unto our F(a) function,
A=XY
 =X(100-2x)
=100-2x²
 =-2x²+100x
F’(a) = 2(-2)x2-1+1(100)x1-1)
              =-4x+100=0
Solve for x we get:
-4x+100=0
-4x100-100=0-100
-4x=-100
(-4x/-4)=(-100/-4)
X=25
Now we plug this back into the original formula
P=2(25)+Y
100=2(*25)+y, 100=50+y, y=50,
Now we know X= 25, and y = 50m so A=XY = (25)(50) = 1,250 sqft

http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx
I must confess, this is not derivative free, because derivatives make life easier.

From what I gather, you want to optimize an area given certain factors.  Let’s assume you were given a property that bordered a building, you would like to get the most sqft given x amount of fence.

First thing we need to do is turn this into a math problem.  We know two things about squares.  That the area A=LW, and the perimeter = 2L+2W.  given we only have 3 sides to our area in question, we assume that P= 2L+W. let’s assign X and Y instead.  Say we have 100 feet of fencing

P=2X+Y; 100=2X+Y.  let’s solve for Y
100=2X+Y
100-2X=2X+Y-2X
Y=100-2x

Now plug that unto our F(a) function,
A=XY
 =X(100-2x)
=100-2x²
 =-2x²+100x
F’(a) = 2(-2)x2-1+1(100)x1-1)
              =-4x+100=0
Solve for x we get:
-4x+100=0
-4x100-100=0-100
-4x=-100
(-4x/-4)=(-100/-4)
X=25
Now we plug this back into the original formula
P=2(25)+Y
100=2(*25)+y, 100=50+y, y=50,
Now we know X= 25, and y = 50m so A=XY = (25)(50) = 1,250 sqft

http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx