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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Rihanna Gamble · Accepted Answer

A line through (0, 2) and (1, 1) would have the smallest area if the given point was (1, 1). (2, 0). Stretch this answer both vertically and horizontally (by a factor of 3) after that (by a factor of 5).Let&#039;s verify using a little calculus and algebra.Suppose the line passes through (t, 0) and (3, 5), where t&amp;gt;3.Then the y intercept is at (0,5tt−3)Consequently, the triangle&#039;s area is:12t⋅5tt−3=5t22(t−3)Therefore:ddt5t22(t−3)=5tt−3−5t22(t−3)2ddt5t22(t−3)=5t2(t−3)2(2(t−3)−t)ddt5t22(t−3)=5t2(t−3)2(t−6)Since we require t&amp;gt;3, the zero of the derivative that we see is when t=6, confirming the geometric solution proposed above.The equation for this line is as follows:5x+3y−30=0

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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