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

1 Answer
Steve M
Nov 23, 2016

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)

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