Find the maximum value of #2x + 2sqrt{x(1-x)}# when #0 \leq x \leq 1#?
Original syntax:
Find the maximum value of #2x + 2sqrt{x(1-x)}# when #0 \leq x \leq 1# ?
Original syntax:
Find the maximum value of
1 Answer
Nov 10, 2017
Explanation:
Let:
#f(x) = 2x+2sqrt(x(1-x))#
#color(white)(f(x)) = 2x+2sqrt(x-x^2)#
Then:
#f'(x) = 2+1/sqrt(x(1-x)) * (1-2x)#
#color(white)(f'(x)) = (2sqrt(x(1-x))+1-2x)/sqrt(x(1-x))#
The numerator of
#2sqrt(x(1-x)) = 2x-1#
Squaring both sides, this becomes:
#4(x(1-x)) = 4x^2-4x+1#
That is:
#4x-4x^2 = 4x^2-4x+1#
So:
#0 = 8x^2-8x+1 = 2(2x-1)^2-1#
So:
#(2x-1)^2 = 1/2#
So:
#2x-1 = +-sqrt(2)/2#
So:
#2x = 1+-sqrt(2)/2#
So:
#x = 1/2+-sqrt(2)/4#
We find:
#f(0) = 0#
#f(1/2-sqrt(2)/4) = 1#
#f(1/2+sqrt(2)/4) = sqrt(2)+1#
#f(1) = 2#
So the maximum value of
graph{2x+2sqrt(x(1-x)) [-0.2, 1.2, -0.4, 2.6]}