Make a sketch.


You have a right triangle ABC with the right angle at vertex A = (0,0); 
the point B is x-intercept (4,0) and the point C is y-intercept (0,3).


So the legs of the triangle are  AB = 4, BC = 3 and the hypotenuse BC is 5 units long.


Let P = (x,y) be the point inside triangle ABC (including its sides).


Then the distance from P to x-axis is y; the distance from P to y-axis is x.



To find the distance from P to the hypotenuse BC, consider triangles PAB, PAC and PBC.

We will find the distance from P to BC as the altitude of the triangle PBC.



The area of the triangle ABC is   =  = 6 square units;

the area of the triangle PAB is   =  = 2y;

the area of the triangle PBC is   =  = 1.5x.

The area of the triangle PAC,  ,  is the difference  -  -  = 6 - 2y - 1.5x.

The altitude of the triangle PBC drawn to BC is TWICE its area divided by the length of BC, i.e.   = 2.4 - 0.6x - 0.8y.


Thus, the sum of the distances from the point P(xy) to the sides of the triangle ABC is  x + y + 2.4 - 0.6x - 0.8y = 2.4 + 0.4x + 0.2y.



The problem asks to find the points P(x,y) inside the triangle ABC, which provide the minimum / the maximum to this function f(x,y) = 2.4 + 0.4x + 0.2y.


This function is linear function of x and y, and it is not a constant.


THEREFORE, the function gets its maximum/minimum inside the triangle at one of the vertices of the triangle.


So, I prepared the Table below, which shows the value of the function at the vertices of the triangle ABC



            T    A    B    L    E


                          f(x,y)

    Point A (0,0)      2.4 + 0.4*0 + 0.2*0 = 2.4

    Point B (4,0)      2.4 + 0.4*4 + 0.2*0 = 4.0

    Point C (0,3)      2.4 + 0.4*0 + 0.2*3 = 3.0



From the Table,  the sum of the distances is maximum at the vertex B(4,0).

                 The sum of the distances is minimum at the vertex A(0,0).