How do I find the maximum and minimum of a function in a given domain?
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I'm trying to find the max and min of a function over a function, but I can't seem to figure out how. My equation is y = (1*x^4)/4+(4*x^3)/3- 5*(x^2)/2 over -3<=x<=3. I tried doing min(y) and max(y) but it's not working. Does anybody know how to find the max and min???
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Image Analyst
2021년 7월 23일
Other than editing away most of your question, what else have you done? Did you like any of the Answers below?
Rik
2021년 7월 23일
Original post (in case Ria decides to edit it away again):
How do I find the maximum and minimum of a function in a given domain?
I'm trying to find the max and min of a function over a function, but I can't seem to figure out how. My equation is y = (1*x^4)/4+(4*x^3)/3- 5*(x^2)/2 over -3<=x<=3. I tried doing min(y) and max(y) but it's not working. Does anybody know how to find the max and min???
답변 (3개)
Rik
2021년 7월 18일
You need a function like fminbnd:
y =@(x) (1*x.^4)/4+(4*x.^3)/3- 5*(x.^2)/2;
x_min = fminbnd(y,-3,3)
Let's confirm this with a plot:
fplot(y,[-3 3])
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Image Analyst
2021년 7월 18일
Try this:
x = linspace(-3, 3, 1000);
y = (1*x.^4)/4+(4*x.^3)/3- 5*(x.^2)/2;
plot(x, y, 'b-', 'LineWidth', 2);
grid on;
% Find where min is
[yMin, indexOfMin] = min(y);
fprintf('Min of y at x = %f, y = %f.\n', x(indexOfMin), min(y));
You get
Min of y at x = -3.000000, y = -38.250000.
Is that what you were looking for?
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Walter Roberson
2021년 7월 19일
syms x
y = (1*x.^4)/4+(4*x.^3)/3- 5*(x.^2)/2
LB = -3; UB = 3;
xcrit = solve(diff(y, x),x)
xcrit(xcrit < LB | xcrit > UB) = [];
xcrit = unique([xcrit; LB; UB])
ycrit = subs(y,x,xcrit)
[miny, minidx] = min(ycrit)
[maxy, maxidx] = max(ycrit)
fprintf('minimum is %g at %g\n', miny, xcrit(minidx))
fprintf('maximum is %g at %g\n', maxy, xcrit(maxidx))
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