Max and Min Calculus Problem?

Question

A right triangle is formed in the 1st quadrant by the x and y axis and a line through the point (2,3). Find the vertices of the triangle so that its area is a minimum.

Answer 1

cv

Answer 2

A line through the point (2,3) has an equation:

y-3=k(x-2) or y=kx+3-2k

y-intersept: x=0 ==> y=n=3-2k

x-intersept: y=0 ==> x=m=(2k-3)/k=2-3/k

Area of the triangle:

A=0.5*m*n

If m=n then A=Amax

2-3/k=3-2k

2k-1-3/k=0

2k²-k-3=0 ==> k1=-1 or k2=3/2

k2=3/2 is not a solution, because in this case the line passes

through the coordinate origin ==>

k=-1, m=n=5, Amax=0.5*5²=12.5