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

Question

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

Answer 1

Hi Daitong,

The line in the first quadrant through (3,5) and forming a triangle has an x-intercept of (x,0) and a y-intercept of (0,y). The slope of that line is -y/x. Use the slope-intercept form of the equation of a line with the point (3,5) and the slope -y/x to get 5=(-y/x)3 + y since the y-intercept is y. Solve for y and get  y = 5x/(x-3). The area of a triangle is A = 1/2 bh. Substituting the base of the triangle which is x and the height which is 

y = 5x/(x-3) into the area formula for a triangle we get

                  A = (1/2)x(5x/x-3) = (1/2)(5x²/x-3).  Use the quotient rule to differentiate and get

                  A^' = (1/2)(5x² - 30x/(x-3)²).   Set the first derivative equal to 0 and solve for x resulting in two values, 0 and 30. Obviously x=0 can not produce the maximum area, use x = 30. Now use the points (30,0) and (3,5) to write the equation of the desired line as y=(-1/5)x+28/5