How do you find the critical points and local max and min for y=4x-x^2y=4xx2?

1 Answer
Nov 23, 2016

Maximum at (2,4)(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) y=4xx2=x(4x)
If y=0 => x(4-x) =0 y=0x(4x)=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)