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???

[Image Analyst]

Other than editing away most of your question, what else have you done? Did you like any of the Answers below?

[Rik]

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???

Answer 1

[Rik]

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)

x_min = -2.9999

Let's confirm this with a plot:

fplot(y,[-3 3])

Answer 2

[Image Analyst]

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?

Answer 3

[Walter Roberson]

syms x
y = (1*x.^4)/4+(4*x.^3)/3- 5*(x.^2)/2

y = 

LB = -3; UB = 3;
xcrit = solve(diff(y, x),x)

xcrit = 

xcrit(xcrit < LB | xcrit > UB) = [];
xcrit = unique([xcrit; LB; UB])

xcrit = 

ycrit = subs(y,x,xcrit)

ycrit = 

[miny, minidx] = min(ycrit)

miny = 

minidx = 1

[maxy, maxidx] = max(ycrit)

maxy = 

maxidx = 4

fprintf('minimum is %g at %g\n', miny, xcrit(minidx))

minimum is -38.25 at -3

fprintf('maximum is %g at %g\n', maxy, xcrit(maxidx))

maximum is 33.75 at 3