Let be a function that is bounded by

I am to find the absolute max/min values.

Completely lost! This is significantly more difficult that any examples we went through in class.

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- October 15th 2010, 10:09 PMEm Yeu AnhTough Max/Min Problem
Let be a function that is bounded by

I am to find the absolute max/min values.

Completely lost! This is significantly more difficult that any examples we went through in class. - October 16th 2010, 04:39 AMHallsofIvy
You appeared to have copied the problem wrong. . No matter what t is, x is never less than -1 and, in particular, is never equal to -3. Since the "boudaries" given do not intersect, they cannot bound a region.

- October 16th 2010, 09:49 AMEm Yeu Anh
- October 17th 2010, 06:12 AMHallsofIvy
Mathematics requires great precision. If you do not take the time and care to

**read**the problem correctly you are are never going to get anywhere. " " has exactly the same difficulty as " ". so that x is never less than -1. The curve given by , does not intersect x= -3 so the curves you give do NOT bound a region.

If it were , when or , , t= -1 and t= 3. At those points, y= 1 and 9.

Now, find the gradient of F(x,y) and set it equal to 0 to find any points in the**interior**of the region at which there**may**be max or min.

Then set x=-3 in F(x,y) to get a function in the single variable y giving the value of F on the line x= -3. You can set the derivative of that equal to 0 to find any critical points there (remember that y must be between 1 and 9 to be on that boundary).

Set , in F to get a function of the single variable t on the parabola. Set its derivative equal to 0 to find any critical points there (remember that t must be between -1 and 3).

Evaluate the function F(x,y) at each of the critical points you have found, as well as at the vertices (-3, 1) and (-3, 9) to determine the largest and smallest values of F in that region.