Doug C. answered • 07/21/21
Math Tutor with Reputation to make difficult concepts understandable
The equation of the line starts out looking like this:
y-3=m(x-4) , point slope. We want to find a value for m that minimizes the area of the triangle in the first quadrant. In order for that triangle to exist m < 0, otherwise the line will not intersect both the positive x and y axes.
To find the y-intercept let x = 0, resulting in y=-4m+3, so the y-intercept is represented by (0, -4m+3).
To find the x-intercept let y = 0, resulting in x = (4m-3)/m, so the x-intercept is ((4m-3)/m, 0).
Then a function for the area of the triangle is A = 1/2 ((4m-3)/m) (-4m+3).
After simplification that results in:
A = -8m -9/2 m-1+12
Find A', set equal to zero to find the critical numbers, prove that that critical number creates a minimum area (first derivative test). Once you have the critical number that creates the minimum area, substitute for m into the point-slope equation of the line. For example if m as a critical number turns out to be -0.8 (it's not), then the equation of the line would be y - 3 = -4/5(x-4).
Use this graph to confirm your work:
desmos.com/calculator/agmaa5dumy
