Tough Max/Min Problem

**Em Yeu Anh** · October 15th 2010, 10:09 PM

Let $F(x,y) = 2(1-x)^{\frac{1}{2}}+(x+y)^2+x-2y$ be a function that is bounded by $x=-3, y(t)=t^2, x(t)=2t+t^2$
I am to find the absolute max/min values.
Completely lost! This is significantly more difficult that any examples we went through in class.

**HallsofIvy** · October 16th 2010, 04:39 AM

You appeared to have copied the problem wrong. $x= t^2+ 2t= t^2+ 2t+ 1- 1= (t+1)^2- 1$ . 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.

**Em Yeu Anh** · October 16th 2010, 09:49 AM

Originally Posted by HallsofIvy

You appeared to have copied the problem wrong. $x= t^2+ 2t= t^2+ 2t+ 1- 1= (t+1)^2- 1$ . 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.

Sorry, should be $x=t^2-2t.$

**HallsofIvy** · October 17th 2010, 06:12 AM

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. " $x= t^2- 2t$ " has exactly the same difficulty as " $x= t^2+ 2t$ ". $x= t^2- 2t= t^2- 2t+ 1- 1= (t- 1)^2- 1$ so that x is never less than -1. The curve given by $x= t^2- 2t$ , $y= t^2$ does not intersect x= -3 so the curves you give do NOT bound a region.

If it were $x= 2t- t^2= 1- (1- 2t+ t^2)= 1- (t- 1)^2$ , $x= -3$ when $-3= 1- (t-1)^2$ or $(t-1)^2= 4$ , $t- 1= \pm 2$ , 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 $x= 2t- t^2$ , $y= t^2$ 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.

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