This function has a closed interval of (-4, 0), (4, 0), while passing through the origin.

I'm struggling to find the maxima and minima of the function, since this finction doesn't have a standalone constant. According to my book, f has a relative minimum if f'(x) changes from negative to positive at (c, f(c)), and a relative maximum from positive to negative at (c, f(c)).

I calculated the first derivative as $\sqrt{16-x^{2}} -\frac{x^{2}}{\sqrt{16-x^{2}}}$ and the critical points are at $x= -4, 0, 4$. I've been taught that to find the relative max/min by plugging in the critical number(s) into f(x), but in this case, zero is the only output.

One other difficulty I have is these calculations require a number from a given interval. Other than picking a number at random or testing each possiblilty, how do you find the min/max when there is no constant number in the function?

  • $\begingroup$ Are you sure you've found the correct critical points? I find them to be $x=-4,\pm 2\sqrt{2}, 4$ based on your first derivative. $\endgroup$ – yunone Apr 1 '11 at 19:43
  • $\begingroup$ I wouldn't be surprised if that was it. finding the critical points is my weak spot right now. $\endgroup$ – Jason Apr 1 '11 at 19:53
  • $\begingroup$ Yes, I think that may have been the snag. I've just posted an answer showing how to calculate them in this case. Hopefully it helps. $\endgroup$ – yunone Apr 1 '11 at 19:54

I believe you have found the incorrect critical points. To find them, you set $f'(x)=0$, in this case, $$ \begin{align*} \sqrt{16-x^{2}} -\frac{x^{2}}{\sqrt{16-x^{2}}}=0 &\implies \sqrt{16-x^2}=\frac{x^{2}}{\sqrt{16-x^{2}}} \\ &\implies 16-x^2=x^2 \\ &\implies x^2=8 \\ &\implies x=\pm 2\sqrt{2} \end{align*} $$ So your critical points are $x=-4,\pm 2\sqrt{2},4$, when you include the endpoints of your interval. Plugging back in, you find $f(4)=f(-4)=0$, and $f(2\sqrt{2})=8$ and $f(-2\sqrt{2})=-8$. From this you can conclude what the relative extrema are on your interval.

  • $\begingroup$ Thats the part I was missing. $\endgroup$ – Jason Apr 1 '11 at 20:00

Hint: $\sqrt{16 - x^2}$ is just $y$ in a circle with radius 4, so you're really maximizing the area of a rectangle inscribed in a circle. What shape would that be?

As far as the calculus goes, I think you took the derivative correctly and messed up with the algebra in finding critical points.


Where did you find a critical point of 0? You'd like $f'(x)$ to be 0 or undefined.


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