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Can I solve a constrained optimization of general form of algebraic equation using matlab?

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Tanay Agarwal
Tanay Agarwal 2021년 7월 5일
답변: John D'Errico 2021년 7월 5일
I'm working with a fairly complex equation and I need to carry out constrained optimization of the same. The first order differential equations are very messy to solve by hand and hence I thought to use online calculators. Problem with calculators like wolfram or even matlab functions I know is that I cannot choose the variables with respect to which optimization needs to be done. If I use specific numbers in place of other variables it would defy the purpose. Can matlab be used for the problem I have?
For example:
Maximize a*x^b x
w.r.t x
s.t. x>0 , a>1 , 0<b<1
Answer here would be in form of a,b. The original problem is multivariable and much more complex. I don't want to optimize with specific values for a,b.

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John D'Errico
John D'Errico 2021년 7월 5일
The problem is, even a simple function as you have written is is not as simple as you think. This is a problem of understanding the mathematics, before you just throw it into a tool like MATLAB or Mathematica.
If you want to maximize that function over an interval, then the solution will occur either at a root of the derivative, OR it will be at one of the end points of the interval. here the interval is semi-infinite, thus x>0, so open at the left end and unbounded on the right.
With 0 < b < 1, we will have a derivative singularity at x==0, but even so, at x==0, the function itself will be zero. We can look at an example curve, here with a=2, and b=0.25.
f = @(x,a,b) a*x.^b - x;
fplot(@(x) f(x,2,0.25),[0,10])
That sample curve is actually rather common for the members of this family, with positive a and b in the interval (0,1). All such curves will have the same general shape.
The maximum will therefore lie at the root of the first derivative:
syms a b positive
syms x positive
xmax = solve(diff(a*x^b - x,x,1) == 0)
Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.
xmax = 
Of course, I could have done that by hand as easily. But this is a MATLAB forum.
Sadly, this really does not solve your problem, which is surely vastly different than this trivial example. But my point is, what you need here is to spend some time with the mathematics that would be employed. No, you may not be easily able to just throw it into a general calculating tool of any sort. Of course, I cannot know, since we have not seen your real problem.

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