An isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the axis on the curve y=27-x^2. find the largest area the triangle can have.

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accessdenied · Answer

Well, we are trying to place the vertices on y=27-x^2.  The graph of y is just an upside-down parabola 27 units above the origin, so this should work out well in terms of our points.  First, we know that the vertices should be the same height (just so that we do have an isosceles triangle as requested).  We'll choose arbitrary points -x and x for the x values to be equally spaced from the origin, thus giving us the desired isosceles triangle.
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The area of the triangle is given by the formula A=(1/2)bh.  The base will be equal to 2x because our points are both 'x' from the origin.  Because the base is also parallel with the x-axis, we know that y is going to be the height of the isosceles triangle.

$$ A = bh => A = (2x)(27-x^{2}) $$

From here, we're just doing an optimization problem to find the maximum of A.  Would you be able to get it from here?