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Calculus/max- min optimization

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Question
An isosceles triangle is drawn with its vertex at the origin and its base parallel to the x axis. The vertices of the base are on the curve 5y = 25 - x^2. find the area of the largest triangle.

Answer
The area of a triangle is known to be bh/2 where b is the base and
h is the height.  In this triangle, the height is y and the base is
twice the x coordinate, or 2x.  This would mean that bh/w
would be 2xy/2 = xy.

Now from the equation, we know that y = (25 - x²)/5, so the area
is x(25-x²)/5.  This can be rewritten as (25x-x^3)/5.  Taking the
derivative gives us (25 - 3x²)/5.  Set this to 0 and it gives
the equation 25 = 3x², or x²=25/3, or x²=3*25/9, so that the
squareroot can be taken easily, giving x=±5√3/3.  Note that if
x as taken as -5√3/3, the same triangle is generated as if x
were taken as 5√3/3, since they are just the same triangle reversed.

Taking the second derivative will make sure that this is a maximum.
The second derivatve is -2x/5.  This will be negative since we took
x as the positive value (since area is usually thought of as
a positive thing), which means that the point is a maximum.

Now if we took the negative value, it would be a minimum.
That is, the greatest negative area possible, but area is
usually not thought of being negative ... but it could be ...

Calculus

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