Expert: Socrates - 12/1/2009

Question
Can you help please? Minimize the area of a right triangle in quadrant one (nested in the origin).  The hypotenuse passes through the point (1,2). The area is to be expressed in terms of the length of the base.

Answer
Let's suppose the line passes through (t,0) on the x axis. So the base of the triangle has length t.

The line will then have slope (2-0)/(1-t) = 2/(1-t)

The line has equation y = (2/(1-t))x + b , where b is the y intercept.

To find b in terms of t ,  you can plug in x=1 , y=2 , because (1,2) is a point on this line.

2 = (2/(1-t))(1) + b

b = 2 -  2/(1-t)

b = -2t/(1-t)

so the y intercept of the line is 2t/(t-1), and this is the height of the triangle.

The area of the triangle is 1/2 the base times the height , so

A(t) = (1/2)(t)(2t/(t-1))

A(t) = t^2/(t-1)

Find the derivative

A'(t) = (t^2-2t)/(t-1)^2

Set this equal to 0

(t^2-2t)/(t-1)^2 = 0

t^2 - 2t = 0

so t=2 or t=0

We can't have t=0 or there is no triangle , so t=2

The minimum must occur when t = 2

To get the area ,

A(2) = 2^2/(2-1)

A(2) = 4

The minimum area for the triangle is 4 .