Mo logo [home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff] german flag
Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

The tangent line $ g$ to the parabola

$\displaystyle y = x(1-x) $

intersects the positive $ x-\!$axis at point $ A$ and the positive $ y-\!$axis at point $ B$, respectively.


\includegraphics[width=.6\linewidth]{TdM_13_A1_bild}


Find the equation of the tangent line $ g$ as a function of the $ x-\!$coordinate $ t > 1/2$ of the point of tangency $ P$.
For which point $ P_{\min}$ is the area $ F$ of the triangle $ \bigtriangleup (O, A, B)$ minimal, and what is the size of the minimum area $ F_{\min}$?


Answer:

$ g$ for $ t=3/4: y = $ $ x$ +  
$ P_{\min} = $ $ \big($ , $ \big)$  
$ F_{\min} = $  

(The results should be correct to four decimal places.)

Solution:

The point-slope form of the tangent line at the point $ P=(x_0, y_0)$ is

$\displaystyle \dfrac{y - y_0}{x - x_0} = s\ . $

With $ x_0 = t$, $ y_0 = t - t^2$ and $ s = 1-2t$, we obtain

$\displaystyle y - (t - t^2) = (1-2t) (x - t) $

or

$\displaystyle y = (1-2t) x + t^2. $

Setting $ y = 0$ and $ x = 0$, respectively, yields
$ A = \left( \dfrac{t^2}{2t - 1}, 0 \right) $


and


$ B = \left( 0\,,t^2\right)\ . $
Hence, we have
$ F(t) = \dfrac{t^4}{4t - 2}\ , \quad t>\dfrac{1}{2}\ .$

Setting the derivative to zero leads to

$ 0 = F'(t) = \dfrac{4 t^3 (4t - 2) - 4 t^4}{(4t - 2)^2}$ $ \quad \Longleftrightarrow \quad 0 = 12 t^4 - 8 t^3 $
yielding the admissible solution

$\displaystyle t = \dfrac{2}{3}\ . $


Since $ F(t) \rightarrow \infty $ for $ t \rightarrow 1/2 $ and $ t \rightarrow \infty$, we obtain

$ P_{\min} = \left(\dfrac{2}{3} , \dfrac{2}{9}\right) \approx (0.667, 0.222) $


and

$\displaystyle F_{\min} = F\left(\dfrac{2}{3}\right) = \dfrac{(2/3)^4}{8/3 - 2} = \dfrac{8}{27}\ \approx 0.296\ . $

[problem of the week]