Asked by Frank
                Please help with this. I submitted it below and was asked to clarify by Henry (hope this is clearer and thank you). I need the first derivative of f(x)=4(x+ sqrt (x(8-x)).And then the second derivative of this to show by the second derivative test that it is a local maximum. This has been difficult to put into form here.Many thanks.
            
            
        Answers
                    Answered by
            MathMate
            
    First 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.
    
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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