MHB Problem of the Week #115 - June 9th, 2014

[Chris L T521]

Sorry about the delay! I had an emergency I needed to tend to the last couple days... >_>

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: Find the equation of the line through the point $(3,5)$ that cuts off the least area from the first quadrant.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by kaliprasad, lfdahl, magneto, and MarkFL. You can find lfdahl's solution below.

[sp]Let $\alpha$ denote the slope of the straight line, which passes through $(3,5)$ and $(x,0)$:
\[\alpha =-\frac{5}{x-3},\: \: \: \: \: \: \: x > 3.\]

The right triangle defined by the legs: $-\alpha x$ and $x$ has the area function:
\[A(x) = -\frac{\alpha }{2}x^2 = \frac{5}{2}\cdot \frac{x^2}{x-3} \: \: \: \: \: \: \: x > 3.\]

From the expression it is obvious, that $A(x)$ has global minimum, when: $A’(x)=0$:

\[A'(x) = \frac{5}{2}\cdot \left ( \frac{2x}{x-3}-\frac{x^2}{(x-3)^2} \right )=0\\\\ \Rightarrow 2x(x-3)-x^2 = 0 \Rightarrow x = 6\]

Therefore, the equation of the line through (3,5), that cuts off the least area in the first quadrant, is:

\[y = -\frac{5}{3}\cdot x+ 10.\][/sp]