mutiplication and division of a range of number.

it seem my previously question has been worded badly, so i will try again.

let say a b and c is a number from 1 to 10, d is a number from 5 to 20.

so F=(a/(b*c))*tan(d) <---not in matlab format

and i trying to find what a b c and d is when F=max.i tried alot of different way to do it, but it all return with matrix error.

[kevin]

for example

a=[1:1:3] % Row vector using your strange syntax.

b=[3:1:5] % Another row vector.

c1 = a' * b % Note transpose operator '

a =

1 2 3

b =

3 4 5

c1 =

3 4 5

6 8 10

9 12 15

this is fin when when a and b is a 1x3 vector, but what do you do when a=1 2 3 4 and b =2 3 4 5 6 7 8?

Answer 1

[Andrei Bobrov]

variant 1

F=@(x)-x(1)/(x(2)*x(3))*tand(x(4))
out = fmincon(F,[1 1 1 5],[],[],[],[],[1 1 1 5],[10 10 10 20],[],optimset('Algorithm','active-set'))

variant 2

[a b c d] = ndgrid(1:10,1:10,1:10,5:20);
v = [a(:),b(:),c(:),d(:)];
F=@(x)x(:,1)./(x(:,2).*x(:,3)).*tand(x(:,4));
out1 = F([a(:),b(:),c(:),d(:)]);
[n,n] = max(out1);
out2 = v(n,:);

[kevin]

if i use variant 2, how accurate is it? i mean like how many decimal places?

[kevin]

can anyone explain F=@(x)x(:,1)./(x(:,2).*x(:,3)).*tand(x(:,4));?

[Andrei Bobrov]

Hi Kevin! Please read:

1. http://www.mathworks.com/help/techdoc/ref/specialcharacters.html

2. http://www.mathworks.com/help/techdoc/ref/arithmeticoperators.html

3. http://www.mathworks.com/help/techdoc/ref/function_handle.html

4. http://www.mathworks.com/help/techdoc/ref/tand.html

Use in your works MATLAB help.

Answer 2

[Walter Roberson]

(a/(b*c))*tan(d) would be maximum when abs(a) is maximum, abs(b*c) is minimum, and abs(tan(d)) is maximum.

As you know the ranges of the variables and know the sign of them is consistent, you can just plug in the appropriate boundary conditions: maximum a, minimum b and c, maximum d, so that would be a=10, b=1, c=1, d=20, which would give 10/(1*1) * tand(20) which is 10*tand(20), which is a value that is approximately 3.64 .

[kevin]

how would you write out the boundary condition? and the equation it self

[Walter Roberson]

ab = max(a);

bb = min(b);

cb = min(c);

db = max(d);

out = (ab / (bb*cb))*tand(db);

More generally,

[A,B,C,D] = ndgrid([min(a),max(a)], [min(b),max(b)], [min(c),max(c)], [min(d),max(d)]);

out1 = A./(B.*C).*tand(D);

[maxval, idx] = max(out1(:));

fprintf(1,'maximum value was: %g at (%g,%g,%g,%g)\n', maxval, A(idx), B(idx), C(idx), D(idx));

This simple coding cannot be used if any of the subexpressions do not change monotonically over the permitted ranges of the variables. For example it could not be used if any of the subexpressions could lead to a division by 0, or if there was a subexpression of the form (x-3)^2 when x ranged below and above 3 (because the result would decrease to a minimum and then increase again)