I have a max/min prob I have been trying to nut out and just can't seem to see it. Any help woudl be very appreciated.
thanks jacs
Understand that is the total length of the two strips. Thus,Originally Posted by jacs
Call side for .
Then, . (Becasue the area is 9).
But, by pythagorean theorem. Thus,
Thus,
Thus,
But, because .
Thus,
Thus,
This answers the first part.
Define a function and attempt to find its minimum value.
This is done by taking the derivative of
Thus,
Simplify this mess,
For simplicity sake let thus,
Now find the critical points (points where there is no deriviative or when it is zero). Since the derivative exists everywhere we find where it is zero, thus,
Thus,
Solve this quadradic,
Thus,
But thus,
as the only critical point.
Thus,
Thus (squaring),
Thus,
Thus,
Thus, ( )
.
Now we prove that is the minimum. We use the first-derivative test (there is no way I am going to use the second-derivative test )
Take any point to the left of say 1 and find the sign of is is negative-the function is decreasing. Take any point to the right of say 4, and find the sign of which is postive-the function is increasing. Thus, if a function switches from decreasing to increasing in a point that point is a relative minimum. Since the function on the interval is decreasing thus . And because the function is increasing on thus, thus, is minimized at
In these long type of problems it is traditional for me to make a mistake somewhere, thus, if my answer does not agree with your books answer try to search where I made a mistake.
Q.E.D.
The PerfectHacker, the solution is perfect, no errors at all.
I can see where i messed up now, i magically dropped my x² in the denominator in my quotien rule, but even asssuming i didnt mess that up, i never would have recognized the hidden quadratic in the final equation.
and you are so right about not wanitng to use the second derivative test on that one
once again, many thanks
jacs