Question 1183764
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Let (x,y) be a point on the triangular region bounded by the line 3x + 4y = 12 and the coordinate axes. 
Determine the points (x,y) in this region which give the minimum and maximum sums of distances of the point 
from the line 3x + 4y = 12 and from the coordinate axes. Please include explanation. Thank you!
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<pre>
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  {{{A[ABC]}}} = {{{(1/2)*3*4}}} = 6 square units;

the area of the triangle PAB is  {{{A[PAB]}}} = {{{(1/2)*4*y}}} = 2y;

the area of the triangle PBC is  {{{A[PAC]}}} = {{{(1/2)*3*x}}} = 1.5x.

The area of the triangle PAC,  {{{A[PAC]}}},  is the difference {{{A[ABC]}}} - {{{A[PAB]}}} - {{{A[PAC]}}} = 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.  {{{(12-3x-4y)/5}}} = 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).
</pre>

Solved, answered and thoroughly explained.



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I just fixed my error after notice by @greenestamps, and now you see the fixed correct version.


Thanks to @greenestamps !