Consider the function $g(x)=1-x^2$
.For $x>0$, the tangent line to $g(x)$ forms a right triangle with the coordinate axis. Find the point of the curve such that the right triangle has the smallest possible area.
When I try to minimize the derivative all I get is X=0 (assuming I set the problem up right). Any ideas?
Best Answer
EDIT 1: Relabelled function from $f$ to $g$ as the latter was used in problem
EDIT 2: Continuation of problem at behest of asker
EDIT 3: Made humiliating error when evaluating $G(1/\sqrt{3})$
Okay...
You agree that the slope of the line tangent to the point $(a, g(a))$ has slope of $-2a$. So the x-intercept is given by $a+\Delta a = a +\frac{0-g(a)}{g'(a)}$. Now the y-intercept is given by $(0 - (a+\Delta a))\cdot g'(a) = -(a+\Delta a)\cdot g'(a).$ So the area of the triangle is given by $1/2\cdot -g'(a)\cdot (a+\Delta a)^2$. In this problem, the area is $1/2\cdot 2a \cdot (a+\frac{1-a^2}{2a})^2 = \frac{(1+a^2)^2}{4a} \equiv G(a)$. Now your goal is to maximise this function.
Now, a continuation of the problem:
$4\cdot G'(a) = 4\cdot [\frac{(1+a^2)^2}{4a}]' = 3a^2 + 2 - 1/a^2 = 0$. Now if $a=0$, then $G(a)$ is undefined so without loss of generality suppose $a \neq 0$. Then, $3a^4 + 2a^2 - 1 = 0$. Let $z = a^2$ and guess a solution of $z = 1/3$. By syncretic division, we find that $3a^4+2a^2 - 1 = (3a^2 - 1)(a^2+1) = 0$. Hence, $a = 1/\sqrt{3}$ or $a=-1/\sqrt{3}$. We should check how $G(a)$ behaves as $a \rightarrow \infty$, namely $G(a) \rightarrow \infty$. Finally, $G''(a)>0$ and so our values of $a$ are collectively global minima as desired.
We conclude that the area is $G(1/\sqrt{3})=4\sqrt{3}/9$.