Optimize Kite Area: Find x for Max Area

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My job is to maximize the area of the kite so that it will fly better, faster, higher. A kite frame is to be made from six pieces of wood. The four pieces that form its border have been cut the lengths. So one of the top border is 2 cm long and the bottom border is 4 cm. The total border length is 2+2+4+4=12 cm (a) Show that the area of the kite is given by the function A(x) = x*(sqrt16-x^2) + (sqrt4-x^2).
(b) How long should each of the cross-pieces to be maximize the area of the kite?
For question a, please see my work as follows:-
I have express the top portion of the vertical cross piece as Y and the bottom cross piece as Z. The horizontal cross pieces expressed as 2x.

So, Y^2=2^2-x^2 therefore Y=sqrt(4-x^2), and the for Z=sqrt(16-x^2)
Therefore the A(x)=x[sqrt(16-x^2) + sqrt(4-x^2)]
The derivative of the expression d/dx is x(sqrt(16-x^2)+sqrt(4-x^2)
da/dx=-x^2/sqrt(16-x^2)+sqrt(16-x^2)(1)+1(sqrt(4-x^2)-x^2/sqrt(4-x^2)
From this derivative tried to find x but I couldn't get.
My last simplified was this=-x^2sqrt(4-x^2)+(16-x^2)*sqrt(4-x^2)+1(4-x^2)*sqrt(16-x^2)-x^2*sqrt(16-x^2)
Once I know the value of x, I can solve the 2nd part (b) by my self.
 
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Thats what I get when I shove the last line into some software. So I am guessing that you have went wrong somewhere... Unfotunately its too late at night for me to think clearly. This is just a heads up.
 
awladdurrah said:
My job is to maximize the area of the kite so that it will fly better, faster, higher. A kite frame is to be made from six pieces of wood. The four pieces that form its border have been cut the lengths. So one of the top border is 2 cm long and the bottom border is 4 cm. The total border length is 2+2+4+4=12 cm (a) Show that the area of the kite is given by the function A(x) = x*(sqrt16-x^2) + (sqrt4-x^2).
(b) How long should each of the cross-pieces to be maximize the area of the kite?
For question a, please see my work as follows:-
I have express the top portion of the vertical cross piece as Y and the bottom cross piece as Z. The horizontal cross pieces expressed as 2x.

So, Y^2=2^2-x^2 therefore Y=sqrt(4-x^2), and the for Z=sqrt(16-x^2)
Therefore the A(x)=x[sqrt(16-x^2) + sqrt(4-x^2)]
The derivative of the expression d/dx is x(sqrt(16-x^2)+sqrt(4-x^2)
da/dx=-x^2/sqrt(16-x^2)+sqrt(16-x^2)(1)+1(sqrt(4-x^2)-x^2/sqrt(4-x^2)
From this derivative tried to find x but I couldn't get.
My last simplified was this=-x^2sqrt(4-x^2)+(16-x^2)*sqrt(4-x^2)+1(4-x^2)*sqrt(16-x^2)-x^2*sqrt(16-x^2)
Once I know the value of x, I can solve the 2nd part (b) by my self.

Your expression for dA/dx looks fine, but I think you went wrong when you tried to simplify it.
dA/dx = -x^2 (16 - x^2)^{-1/2} + (16 - x^2)^{1/2} - x^2 (4 - x^2)^{-1/2} + (4 - x^2)^{1/2}

From the first two expressions, pull out a factor of (16 - x^2)^(-1/2). From the last two expressions, pull out a factor of (4 - x^2)^(-1/2). You should get
dA/dx = (16 - x^2)^{-1/2}(-x^2 + 16 - x^2) + (4 - x^2)^{-1/2}(-x^2 + 4 - x^2)
= \frac{16 - 2x^2}{\sqrt{16 - x^2}} + \frac{4 - 2x^2}{\sqrt{4 - x^2}}

Set this last expression to zero and solve for x.
 
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