Optimization problem? Hi I was having trouble figuring out this question. 
Find the point on the circle $x^2 + y^2 = 1$ in the first quadrant where the tangent line to the circle encloses with the coordinate axes a triangle of the smallest area. Justify your answer.
Thank you 
 A: Let the point of tangency be $P=(a,b)$. The tangent line is perpendicular to $OP$, and hence has slope $-\frac{a}{b}$. We find that the tangent line has equation $by+ax=a^2+b^2=1$. The intercepts are at $\frac{1}{a}$ and $\frac{1}{b}$, so we want to minimize $\frac{1}{2ab}$ or equivalently to maximize $2ab$.
From $(a-b)^2\ge 0$ we find that $2ab\le a^2+b^2=1$, with equality if and only if $a=b$. Thus the required point is $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$.
A: if we parametrize the circle as $\{P_{\theta}=(\cos \theta, \sin \theta)\}_{\theta \in [0,2\pi]}$ then the tangent at $P_{\theta}$ has slope $-\cot \theta$ so its equation is
$$
y-\sin \theta = -\cot \theta (x-\cos \theta)
$$
i.e.
$$
x \cos \theta + y \sin \theta = 1
$$
this intersects the $y$-axis at $(0, \frac1{\sin \theta})$ and the $x$-axis at $(\frac1{\cos \theta},0)$
the area of the triangle is thus $\frac1{\sin 2 \theta}$
for $\theta \in [0, \frac{\pi}2]$ this has a unique minimum at $\theta=\frac{\pi}4$ and the required point is $P_{\frac{\pi}4}=(\frac{\sqrt{2}}2,\frac{\sqrt{2}}2)$
A: The simplest way to compute this would be to create a function that takes the size as a function of angle $\theta$ as you move the radius around the circle.
Then you note the point of intersection along the circle's circumference, which is $(cos \theta, sin \theta)$. You then need to calculate the slope of the tangent line as a function of angle. The height of the triangle will then be $$sin \theta +m cos \theta$$
The width of the triangle will be $$cos \theta + {sin \theta \over m}$$
You can then use the geometric formula for the area of the triangle, find the critical number, and use that to find the minimum. You can then calculate the point of contact again as $(cos \theta, sin \theta)$. Note that I didn't calculate for you the actual slope of the tangent line, but it should not be difficult to find.
