Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

enter image description here

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

share|cite|improve this question
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
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 $G(1/\sqrt{3})$


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

share|cite|improve this answer
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.

share|cite|improve this answer
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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.