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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.

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When I try to minimize the derivative all I get is X=0 (assuming I set the problem up right). Any ideas?

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You don't have the correct triangle. The hypotenuse of the triangle should be tangent to the curve $1 - x^2$. –  user61527 Nov 30 '13 at 0:07
    
So doesn't that give us 1/2(1-x^2)(1-x^2)? Maximizing that still gives me X=0 –  user104827 Nov 30 '13 at 0:19
    
Where are the parentheses in that last expression? What is in the numerator, what in the denominator? –  Ross Millikan Nov 30 '13 at 0:24
    
(1/2)(1-x^2)(1-x^2) –  user104827 Nov 30 '13 at 0:28
    
What is the area of the triangle you want to minimize? Well... the area of the triangle is $1/2\cdot 2a\cdot (\frac{1-a^2}{2a}+a)^2$. So... –  Chris K Nov 30 '13 at 0:54

3 Answers 3

up vote 0 down vote accepted

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 $G(1/\sqrt{3})=4\sqrt{3}/9$.

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Where are you getting the change in A from? –  user104827 Nov 30 '13 at 1:20
    
No not really.. –  user104827 Nov 30 '13 at 1:25
    
Hmm... let's say we have a line of let's say slope -2. Then if we move 1 unit in the positive x-direction, we are also moving 2 units in the negative y-direction. We can also write $y(x) = mx$ and so by back substitution $y(x+\Delta x) = m\cdot (x + \Delta x)$. Now, $\Delta y = y(x+\Delta x) - y(x) = m\cdot (x + \Delta x) - mx = m\cdot \Delta x$. Here, $y$ is a function of $x$. –  Chris K Nov 30 '13 at 1:28
    
Well my goal is to minimize the function, and when I do that I am left with X=4, X=-12 and X=-1/3? –  user104827 Nov 30 '13 at 1:50
    
I noticed you made an error at first glance, since the function is symmetric about the $y$-axis and so you should get the same solution for $a$ just with opposite sign. @user104827, better? –  Chris K Nov 30 '13 at 2:03

First calculate the derivative of $g(x)$ this would give $g'(x)=-2x$. From here we will look at the line through a point $(x_0,y_0)$ on $g$ and calculate the formula of the line. If we have the line, then we can caulculate where it intersects the $x$-axis and $y$-axis. And so we can calculate the area of the triangle in terms of $x$. Minimizing this function would give the required answer.

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I do not understand which points we use to calculate the formula of the line. –  user104827 Nov 30 '13 at 0:35
    
The point $(x,g(x))$, and we look at the tangent of this point, to find the line. –  user112167 Nov 30 '13 at 0:41
    
Wouldn't that still leave us with g'(x)= -2x? –  user104827 Nov 30 '13 at 1:04
    
@user104827, yes this gives a derivative of $-2x$. You still have to construct the area as a function of the ordinate at which you choose to take the tangent. –  Chris K Nov 30 '13 at 1:07

Your triangle is wrong. The triangle you're looking for is the red one in the picture below.
The triangle thou seekest.

The equation of the tangent line at $x=a$ is: $$y - g(a) = (g'(a))(x-a)$$

The top of the triangle has the $y$ you get when you plug in $x=0$ to the equation above (that is, $a\ne 0$, but $x$ (the variable) is equal to $0$).

Does that help?

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