Minimize the area of the triangle.

Question

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

Answer 1

Kenneth started well.  b = 3 - 2m; a = 2 - 3/m

Minimize 1/2 ab wrt m; notice that m must be negative for the problem to make sense;

Instead of minimizing 1/2 ab, let's minimize ab; the m value is the same

 

(3 - 2m)(2 - 3/m) = 6 - 4m - 9/m + 6 = 12 - 4m - 9/m

 

We solve this by taking the derivative; -4 + 9/m^2 = 0

4 = 9/m^2

4m^2 = 9

m = +- 3/2; as we have already said, the solution is when m is negative, so m = -3/2

Taking the second derivative, we get -9/m^3; as m is negative, -9/m^3 is positive, so we are at a minimum.

Answer 2

Let the hypotenuse of the right triangle pass through (2,3) and have y-intercept b and x intercept a.

The line containing the hypotenuse has equation y-3 = m(x-2).

m = -b/a

 

(0,b) is on the hypotenuse line so b-3 = m(0-2)

(a,0) is on the hypotenuse line so 0-3 = m(a-2)

 

you solve each of the two equation for a, as a function of slope m.

 

Now the area is ½•ab so in order to maximize area, maximize f(m) = ab by finding the critical point, etc. (Calculus methods)