Question

How do you find the critical points and local max and min for `y=4x-x^2`?

Answer 1

Maximum at `(2,4)`

Explanation:

Although we can use calculus sometimes an intuitive common sense approach is quicker and easier.

`y=4x-x^2 = x(4-x)`
If `y=0 => x(4-x) =0`
`:. x=0, x=4`

The function has a -ve coefficient of `x^2` so it will be `nn` shaped. Therefore it will have a maximum and this will occur at the midpoint of the two roots, ie at `x=2`

`x=2 => y=8-4=4`, so `(2,4)` is a maximum

graph{4x-x^2 [-10, 10, -5, 5]}

Using Calculus

`y = 4x-x^2`
`dy/dx = 4-2x`

At critical point, `dy/dx=0 => 4-2x=0`
`:. 2x=4`
`:. x=2` (as above)

To establish min or max we look at the 2nd derivative:
`dy/dx = 4-2x`
`:. (d^2y)/(dx)^2 = -2`
When `x=2 => (d^2y)/(dx)^2 < 0 =>` maximum (as above)