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April 1, 2015

April 1, 2015

Posted by **Frank** on Friday, March 4, 2011 at 1:48pm.

- calculus -
**MathMate**, Sunday, March 6, 2011 at 7:14amFirst we calculate

f'(x)=4*((8-2*x)/(2*sqrt((8-x)*x))+1)

Equate it to zero and solve for x

f(x)=0 =>

((8-2x)/(2sqrt((8-x)x))+1)=0

(8-2x)/(2sqrt((8-x)x))=-1

(8-2x)=-2sqrt((8-x)x)

Square both sides:

(8-2x)^2=4(8-x)*x)

Solve the quadratic to get

x= 4±2√2.

Verify both roots (because we squared) and reject x=4-2√2

That leaves us with

x=4+2√2.

Calculate f"(x):

-(64sqrt(8x-x²))/(x^4-16x^3+64x^2)

Verify that f"(x) <0 at x=4+√2 and hence a maximum.

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