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If I have a problem such as this:

We need to enclose a field with a fence. We have 500m of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area.

This is obviously a maximizing question, however what would I do differently if I needed to minimize the area? Whenever I do these problems I just take the derivative, make it equal to zero and find x, and the answer always seems to appear - but what does one do differently to find the minimum or maximum?

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The Extreme Value Theorem guarantees that a continuous function on a finite closed interval has both a maximum and a minimum, and that the maximum and the minimum are each at either a critical point, or at one of the endpoints of the interval.

When trying to find the maximum or minimum of a continuous function on a finite closed interval, you take the derivative and set it to zero to find the stationary points. These are one kind of critical points. The other kind of critical points are the points in the domain at which the derivative is not defined.

The usual method to find the extreme you are looking for (whether it is a maximum or a minimum) is to determine whether you have a continuous function on a finite closed interval; if this is the case, then you take the derivative. Then you determine the points in the domain where the derivative is not defined; then the stationary points (points where the derivative is $0$). And then you plug in all of these points, and the endpoints of the interval, into the function, and you look at the values. The largest value you get is the maximum, the smallest value you get is the minimum.

The procedure is the same whether you are looking for the maximum or for the minimum. But if you are not regularly checking the endpoints, you will not always get the right answer, because the maximum (or the minimum) could be at one of the endpoints.

(In the case you are looking for, evaluating at the endpoints gives an area of $0$, so that's the minimum).

(If the domain is not finite and closed, things get more complicated. Often, considering the limit as you approach the endpoints (or the variable goes to $\infty$ or $-\infty$, whichever is appropriate) gives you information which, when combined with information about local extremes of the function (found also by using critical points and the first or second derivative tests), will let you determine whether you have local extremes or not. )

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When you take the derivative of a function $f(x)$ and solve for where it is zero you are finding the points at which the function does not change. This includes maxima and minima. The fact that it always seems to work for you leads me to believe that you don't really know how it works.

When you find the extreme points you need to perform the second derivative or some other test to make sure that what you get is a maximum or a minimum.

So, my answer is that in both cases you do basically the same thing, except in the end where you pick your maximum or minimum. For a concrete example, consider the function $$f(x) = 2x^3-3x^2.$$ We shall find the local minimum and local maximum. When you take the derivative of $f(x)$ you obtain $$f'(x) = 6x^2-6x = 6x(x-1).$$ You see that the function $f(x)$ has two extreme points - one at $x=0$ and one at $x=1$. We take the second derivative to perform the second derivative test $$f''(x) = 12x-6.$$ We see that $f''(0) = -6 <0$ so the function $f(x)$ is concave down, meaning $f(0)$ is a maximum. Also $f''(1) = 6 > 0$, so the function is concave up, meaning that $f(1)$ is a minimum. Note that if $f''(x) = 0$ at some point, the second derivative test gives us no information.

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I can't see how this works for the problem I listed though? I can't seem to find the two stationery points? – Parachuting Panda Nov 6 '11 at 9:30
@ParachutingPanda Your particular function should be $x(500-x)$. Since it's derivative is a first degree polynomial, it has only one stationary point at most. So, you'll check that point, as well as the endpoints of the reasonable domain. – Austin Mohr Nov 6 '11 at 18:23
I just gave this example because it had two stationary points and it was easy to demonstrate that the approach is pretty much the same. As Arturo said, you also need to check your endpoints. – Aleks Vlasev Nov 6 '11 at 23:05

The problem of minimizing and the problem of maximizing are essentially the same.

As you say, take the derivative of your function and find the critical points. Do not forget, however, that global extrema can also occur at endpoints of the reasonable domain (i.e. the values of $x$ allowed within the context).

For the fence problem, suppose we let $x$ represent the width of the fence. You would first write a function for area in terms of $x$. Notice that this function, although technically defined for all reals, really only makes sense on $0 \leq x \leq 500$ , as the width of the fence cannot be smaller than 0 nor larger than 500. Thus, you need to check not only whatever critical points you find, but also $x = 0$ and $x = 500$. To determine which is the maximum and which is the minimum, you simply compare the values of the function at all these interesting points.

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Suppose you call your area function f

Differentiating it, you will get f', or the slope of f. If you equate this slope to 0, you will find a maximum or minimum (because at the point where a function stops increasing and starts decreasing or vice-versa, the slope must be 0).

Now, if you differentiate the SLOPE of f (f') you will get the slope of the slope of f. If you give it a little thought, you'll see the following 2 cases:

f = maximum so, f is increasing before the maxima. This means that f' just before the maxima is +ve (increasing functions have +vce slope), but at the maxima it is 0 This means that f' is actually decreasing. So, the slope of f' (or f'') is negative

Conclusion: A zero slope and a negative slope-of-slope implies a maxima. A zero slope and a positive slope-of-slope implies a minima.

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Assume the field is rectangular.

If the building has W= 100 m width, then breadth B = (500-100)/2 for a U- or C- shaped fence .. straight away. It does not concern maximum/minimum.

If freedom for shape is assumed, then by Dido's iso-perimetric problem (calculus of variations) you can work out a circular layout of 500 m perimeter length.

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