Finding the minimum value
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Hi.I have a function f(x,y)=2.62./2.62+(x-y).^4, based on the attached figure, the minimum value for this function is 1. I want to know among all of the value for x and y that makes f(x,y)=1, what is the minimum of x-y? I mean how can I find the minimum value of x-y from all of the values that make f(x,y)=1.
Thanks in advance for any help
openfig('xy.fig');
4 个评论
Torsten
2022-6-14
x-y is constantly 0 for all values that make f(x,y) = 1.
Maybe you forgot to set brackets in your function definition ?
f(x,y)=2.62./(2.62+(x-y).^4)
instead of
f(x,y)=2.62./2.62+(x-y).^4
?
M
2022-6-14
Walter Roberson
2022-6-14
编辑：Walter Roberson
2022-6-14
For real x and y, (x-y)^4 is always non-negative. 2.62 plus a non-negative value will always be minimum 2.62 or larger. 2.62 divided by a value that is 2.62 or larger, has a maximum value of 1 and cannot be greater than 1. Imagine for example x=1000 y=0 giving 2.62+1e12 as the denominator, the division is going to give near 1e-11
The plot cannot be for the formula you give. The plot could, however, be for the version of the formula missing the ()
M
2022-6-14
采纳的回答
Image Analyst
2022-6-14
Compute your f matrix, then try this
% Find the overall minimum value of f.
minValue = min(f(:)) % Hopefully 1 but maybe not exactly.
% Find where that min value occurs
[minRow, minCol] = find(f == minValue)
Note if the min occurs in more than one element minRow and minCol will be vectors giving all the places where minValue occurs.
8 个评论
farah bouzidi
2022-6-14
hello @Image Analyst i really need your help in my project since i saw that you helped many people , i am preparing my final year project and i am out of luck in solving my problem please could you contact me i could'nt find a way
Walter Roberson
2022-6-14
Image Analyst cannot be contacted for this purpose. The email address he uses outside of his job is from a service that deletes all messages every few hours; if you did manage to find it, he would never receive the message.
Image Analyst
2022-6-14
编辑：Image Analyst
2022-6-14
@M Not sure what you're asking. You say "at what row and column, what is the value of x and y?" Well x is the column so it's minCol, and y is the row so y is minRow. The matrix value at M(minRow, minColumn) or M(y, x) will be minValue.
Walter Roberson
2022-6-14
mask = x > y;
minf = min(f(mask));
[row, col] = find(f == minf & mask);
sympref(FloatingPointOutput=false);
syms x y
f(x,y) = 2.62./(2.62+(x-y).^4)
dfx = diff(f, x)
critx = solve(dfx, x)
%so every y is a triple critical point for x
dfy = diff(f, y)
crity = solve(dfy, y)
%so every x is a critical point for some y
d2f = simplify(diff(dfx, x))
[N, D] = numden(d2f)
xsaddle = solve(N == 0, x)
vpa(simplify(subs(f, x, xsaddle)))
vpa(simplify(subs(f, x, xsaddle - 1/1000)))
vpa(simplify(subs(f, x, xsaddle + 1/1000)))
So the first two critical point xsaddle are maximums, the second and third critical points are places where greater or lower would lead to imaginary outputs; the last two are again maxima.
fsurf(f, [-2 2 -2 2])
Looks to me as if there is a "crease".
M
2022-6-15
更多回答（2 个）
Walter Roberson
2022-6-14
0 个投票
For the revised formula over real numbers the minimum value is when x and y differ as much as possible, in particular if the difference between them is infinite, which would give you an output of 0.
The maximum value occurs when (x-y)^4 is minimal which occurs when x == y exactly, which gives you 2.62/2.62 == 1
tharun
2023-10-26
0 个投票
Find the minimum value f(x,y) = x3 + y 3 subject to the constraints x + y = 20
3 个评论
tharun
2023-10-26
answer needed
Walter Roberson
2023-10-26
When you have an equality constraint relating x and y, solve the constraint for one of the variables, getting a formula for it in terms of the other variable. Then substitute that formula into the main function. You now have a function of a single variable to be optimized. Proceed by way of calculus.
Minimize a bivariate function 
subject to constrained 
,
,can be transformed into an unconstrained optimization problem by taking the substitution method:
This yields 
,
,and the minimum of the quadratic function can be found at
.From the equality constraint, we can find 
.
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