Smallest area of triangle

Question

Let $$f(x) = 3x^3 - 3x^2 +x - 1/9$$ For any $a\in\mathbb{R}$, let $l$ be the tangent line to the graph of $f$ at the point $(a, f(a))$, and let $M$ and $N$ be the $x$- and $y$-intercepts of $l$ respectively. Let $O$ be the origin.

Find the point(s) $(a,f(a))$ so that triangle $\triangle OMN$ has the smallest area.

Hi, I'm wondering if the area of a triangle in a graph can be zero? I obtained three values of a such that two values of a gives zero while the other is a positive value (1/128)

Since the question wants smallest area (i.e minimum area), does that mean I should choose those two values of a?

Thank you.

Answer 1

Note that $f(x)=3(x-1/3)^3$ and $f'(x)=9(x-1/3)^2$. Therefore $f(1/3)=f'(1/3)=0$and the tangent line $y=f'(a)(x-a)+f(a)$ for $a=1/3$ is $y=0$.

It follows that for $a=1/3$ the points $M,N,0$ are along the same line and the triangle $\triangle OMN$ has area zero.

More generally, $$M=((6a+1)/9,0)\quad\text{and}\quad N=(0,-(6a+1)(3a-1)^2/9).$$ and therefore $$|\triangle OMN|= 2(a+1/6)^2(a-1/3)^2.$$ It follows that the minimum area is attained at $a=1/3$ and $a=-1/6$.

Answer 2

No, the question is about the area of a triangle so, if one or two of your tangent lines go through the origin then there is no triangle.

I would specify the values of $(a) $ where the tangent line passes through the origin and the values where we actually get a triangle.

The real problem is that the area is a continuous function of $(a)$ and the minimum area of zero is attained where $$f(a)=af'(a)$$

That is at $a=1/3$ and at $a=-1/6$ thus there is no positive minimum area.