Find the equation of the line through the point #(3, 5)# that cuts off the least area from the first quadrant?

1 Answer
George C.
Apr 3, 2017

#5x+3y-30 = 0#

Explanation:

I think it will be the line through #(0, 10)# and #(6, 0)# cutting off a triangle of area #30# from Q1.

Why?

If the given point was #(1, 1)# then the minimum area would be for a line through #(0, 2)# and #(2, 0)#. Then stretch this solution horizontally (by a factor of #3#) and vertically (by a factor of #5#).

Let's check using a little algebra and calculus...

Suppose the line passes through #(t, 0)# and #(3, 5)#, where #t > 3#.

Then the #y# intercept is at #(0, (5t)/(t-3))#

So the area of the triangle is:

#1/2 t*(5t)/(t-3) = (5t^2)/(2(t-3))#

Then:

#d/(dt) (5t^2)/(2(t-3)) = (5t)/(t-3) - (5t^2)/(2(t-3)^2)#

#color(white)(d/(dt) (5t^2)/(2(t-3))) = (5t)/(2(t-3)^2)(2(t-3)-t)#

#color(white)(d/(dt) (5t^2)/(2(t-3))) = (5t)/(2(t-3)^2)(t-6)#

Since we require #t > 3#, the zero of the derivative that we see is when #t=6#, confirming the geometric solution proposed above.

We can write the equation of this line as:

#5x+3y-30 = 0#

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