2014/052) Find the equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

 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 passes through (3,5)

now x intercept when x = 0 is y = 5 - 3m

now y intercept when y  = 0 is x= (3m-5)/m

so area of the triangle in 1st quadrant is xy/2

so we need to minimize xy = - (3m - 5)^2/m = (3p+5)^2/p where p = -m and p >0

(3p + 5)^2/p = (3p^(1/2) + 5p^-(1/2))^2 = (3p^(1/2) -5p^-(1/2))^2 + 60

it is lowest when (3p^(1/2) -5p^-(1/2)) = 0 or p = 5/3

so equation of line is y - 5 = -5/3(x-3) or 3y + 5 x = 30
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