The derivative has probably unlimited applications in physics, biology, economics and generally in real life.

But let's see how we can use it to solve simple problems like optimizing the usage of a metal sheet when we want to build a metallic bucket or a jar.

So let's suppose that we have a rectangular metallic sheet of Aluminium of 6 dm^{2}. We want to build a jar that will have a volume as much as possible (i.e. we waste the metallic sheet as little as possible). The jar must also have a bottom and a top/cover like in the image below:

**The question is**: how to cut that piece of metal in order to maximize the volume (or to optimize the usage of the metallic sheet) ?

You can do it by trial and error but perhaps you will never get the theoretical maximum volume.

So let's try to understand little bit what we know (or don't know) and what we have to do:

- we have a 6dm
^{2}rectangular metallic sheet- we don't know its width/height, though

- we have to cut 2 circular sheets for the jar's top and bottom/base
- we don't know the optimal radius, this we'll have to determine

- we have to cut a rectangular sheet for the jar's lateral side
- we don't know the length and the height, yet

**2.a**) Let's assume that the optimal radius should have the value of *x*.

**3.a**) If we want a cylindric jar then the length of the rectangular sheet used for the jar's lateral side should be chosen with respect to the top/bottom radius, i.e. 2πx.

Ok, but how about its height?

Well, if we want to use all the metallic sheet (in practice it's not possible but let's assume that in theory it will work) then we have to see what we have and what we have used so far:

- we've started with 6dm
^{2} - we've cut 2 times (once for the top and once for the base/bottom) an area of πx
^{2} - we still have available an area of 6dm
^{2 }- 2πx^{2}- from this area we have to get a rectangle with the length 2πx and the height equal to what we still have (ie. 6dm
^{2 }- 2πx^{2}) divided by the height (i.e. 2πx); also

- from this area we have to get a rectangle with the length 2πx and the height equal to what we still have (ie. 6dm

So, what's the volume of the cylinder? Well, that's the base area multiplied by the lateral side height, i.e.

For what arbitrary x will we get the maximum volume? Here is where calculus comes once again in rescue.

In order to determine the maximum of a function (eg. maximum of V(x)) we have to determine the x-value where y-value is maximum. Up to that point the function increases its value and then suddenly after that point the curve (the function) starts to decrease its value. Just at that point (x,V(x)) if we draw a tangent to the curve it will be parallel with *x*-axe. The tangent to the curve at a point it's nothing more than the derivative of the original function at that point. So let's derivate the V(x):

We said earlier that the tangent to the curve at a point (x) is parallel with the *x*-axe. So the question is: at what x-value the tangent (V'(x)) is equal with the x-axe (y=0) ?

V'(x)=y=0 <=> V'(x)=0 <=> -3x^{2}π+3=0 => . So when then V(x) will have its maximum value, i.e. we got our maximum volume. Let's plug the x-value in V(x) and see what it gets:

So in order to get the maximum volume possible (i.e. 1,128dm^{3}) we have to cut 2 circles with radius and then a rectangle with the length and the height . If we sum up these areas we'll get exactly 6dm^{2}.

How cool is that?

Now, if you think that this article was interesting don't forget to rate it. It shows me that you care and thus I will continue write about these things.

#### Eugen Mihailescu

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