MaplePrimes - answers and comments on Question, Maximum and Minimum Please help me !

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Markiyan Hirnyk — Tue, 21 Feb 2012 01:39:55 Z

> f := proc (x) options operator, arrow; x*(2011+sqrt(2013-x^2)) end proc:
> plot(f, -sqrt(2013) .. sqrt(2013));

> solve(diff(f(x), x) = 0);


                           -2* 503^(1/2)     , 2 *503 ^(1/2)
> evalf(f(2*sqrt(503)));

                                 90248.90984
> evalf(f(sqrt(2013)));

                                 90226.46825

My way

toandhsp — Tue, 21 Feb 2012 15:22:21 Z

f:=x->x*(2011+sqrt(2013-x^2));

maximize(f(x),x=-sqrt(2013)..sqrt(2013),location);

minimize(f(x),x=-sqrt(2013)..sqrt(2013),location);

Could you do more clearly?

yangtheary — Tue, 21 Feb 2012 03:52:14 Z

Could you do more clearly?

Explanation

Markiyan Hirnyk — Tue, 21 Feb 2012 09:21:29 Z

@yangtheary We find the critical points of the function under consideration, i.e. the roots of its derivative and the points of the domain of f, where the derivative does not exist. Then we determine the change of its sign as follows.

> der := D(f);



> evalf(der(2*503^(1/2)-0.1e-1));

                                 552.219341
> evalf(der(2*503^(1/2)+0.1e-1));

                                -4266.916499

Becouse the derivative changes its sign from + to -, we conclude that x= 2 *503 ^(1/2) is the minimum Edit. maximum point of f. Consider x= -2* 503^(1/2) on your own. See a textbook for further details.

PS. More exactly,