2015/097) Find the equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

Let the slope of line be m

clearly m is -ve as if m is positive or zero then it shall not cut x and y both in 1st quadrant
so the equation of line is

 $y-5 = m(x-3)$


as it goes via $(3,5)$
 
now x intercept when x = 0 is $y=5-3m$

y intercept when y = 0 is given by $x = \dfrac{3m-5}{m}$    
 so area of the triangle in 1st quadrant is $\dfrac{xy}{2}$
so we need to minimize $xy = \dfrac{(3m-5)^2}{m}$

let m = - p where p > 0

$xy=\dfrac{(5+3p)^2}{p}=(\frac{5}{\sqrt{p}}+3\sqrt{p})^2$
or  $xy=(\frac{5}{\sqrt{p}}-3\sqrt{p})^2+60$

clearly it is lowest when  $\frac{5}{\sqrt{p}}-3\sqrt{p}=0$


 so $p=\frac{5}{3}$

so equation of line is
$y=5-\frac{5}{3}(x-3)$
or $3x+5y=30$

Ths problem appeared as Problem of the week in http://mathhelpboards.com/potw-university-students-34/problem-week-115-june-9th-2014-a-10908.html

where you can find some solution. I answered the question correctly but my solution is not provided there.