Enclosing a parabola

Since we know that the congruent sides of the triangle are tangent to the graph of $9- x^2$ , we can use that definition to define the lengths of the base and height of $\Delta ABC$ in terms of the point of tangency.

Let the point of tangency of one side be at $x = k$ . Similarly, it would be $x = -k$ for the other side. We can now get the point-slope form of the tangent line.

$\large y - (9- k^2) = m (x - k)$

where $\large m = \frac{d}{dx}_{x=k} [9 - x^2] = -2k$

So we have

$\large y - (9-k^2) = -2k (x-k)$

$\large y = -2kx + 2k^2 + 9 - x^2$

$\large y = -2kx + k^2 + 9$

This now gives the equation of the tangent line passing through the graph of $9 - x^2$ at $x = k$ for every chosen $k$ .

We'll just use its intercepts though. This is because the y-intercept represents the height, and the x-intercept represents half the base.

$\large y_0 = k^2 + 9$

$\large x_0 = \frac {k^2 + 9} {2k}$

So, now we have an explicit formula for the area of $\Delta ABC$ in terms of the parameter $k$ .

$\large A ( k) = \frac {(k^2 + 9)^2} {2k}$

And from here we can now solve for its extreme values. Notice that as $k$ approaches $0$ , the area increases without bound. Thus it is ruled out that there exists a maximum value, so we look for its minimum. Notice also that our feasible range of values for $k$ run along $-3 \leq k \leq 3$ .

$\large A' (k) = 0$

by the quotient rule,

$\large \frac { 2k(4k^3 + 36k) - 2(k^4 + 18k^2 + 81) }{4k^2} = 0$

$\large 6k^4 + 36k^2 -162 = 0$

$\large k^4 + 6k^2 - 27 =0$

where $\large k = \pm \sqrt {3}, \pm 3i$

So now, we use the fact that $k = \pm \sqrt{3}$ , and we get

$\large A ( k) = \frac {((\sqrt {3})^2 + 9)^2} {2\sqrt {3}}$

$\large = \frac{144}{2\sqrt{3}}$

$\large = 24 \sqrt{3} \approx \boxed{ 41.569}$