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awladdurrah
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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.
(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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