Fenced area - finding maximum by derivatives

 

The area is `A = xy`

We know `2x + y = 800` so `y = 800 − 2x`

So the area is A = x(800 − 2x) = 800x − 2x2

To maximise the area, find when `(dA)/dx = 0`

`(dA)/(dx)=800-4x=0`

when

` x=200`

Is it a maximum?

`(d^2A)/(dx^2)=-4<0\ "for all"\ x`

So it is a maximum.

So the maximum area occurs when `x = 200`, `y = 400` and that area is:

A = 200 × 400 = 80 000 m2 = 8 ha

Easy to understand math videos:
MathTutorDVD.com