How do derivatives provide answers for optimization problems ?

[urbano]

I' at a a very very very basic level of calculus and usually have to watch a video or read something basic just to understand the basics.

I'm fascinated by optimization equations, for example what is the largest area that can made with 500m of fencing. So at some point in solving this we end up using the first derivative.

My limited understanding of first derivatives however relates to finding the slope of of a point on a curve.

What I am currently struggling to understand is how does finding the slope on a point of a curve relate to finding the maximum area that can be created with a certain amount of fencing?

thanks in advance for any help with this

 

[PeroK]

The slope of a curve at a maximum or minimum is zero, hence the derivative of the function is zero at these points.

 

[Jilang]

Consider an area with sides of length a and b so that 2(a+b) =500. The area is given by ab and by substitution a(250-a). If you plot the graph of area (on the y axis) as a function of a (on the x axis) will get a curve:
Y = 250x -x^2 which will increase initially and them start to fall.
You are trying to find the point where the area is a maximum and that will when y is a maximum. The gradient of the curve at that point will be zero.

 

[HallsofIvy]

If the derivative at a point is positive the function value is increasing- increasing x slightly will give a higher value so the point is not a maximum, decreasing x slightly will give a lower value so the point is not a minimum.

If the derivative at a point is negative the function value is decreasing- increasing x slightly will give a lower value so the point is not a minimum, decreasing x slightly will give a higher value so the point is not a maximum.

A maximum or minimum can only occur where the derivative is 0.

But note that this is a "necessary condition", not a "sufficient condition". The derivative of $y= x^3$, $3x^2$, is 0 at x= 0 but there is neither maximum nor minimum there.