Question 1183764
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The response from the other tutor contains errors, which I am sure she will find and correct after seeing that I have provided a response to your question.<br>
The given triangle has vertices (0,0), (4,0), and (0,3).<br>
Let the point P (x,y) be in that triangle.<br>
The distance of P from the x-axis is y; the distance from the y-axis is x.<br>
The distance of P from the line 3x+4y=12, or 3x+4y-12=0, is<br>
{{{abs((3x+4y-12)/sqrt(3^2+4^2))=abs((3x+4y-12)/5)}}}<br>
For all points inside the triangle,<br>
{{{(3x+4y-12)/5)}}}
is negative.  So<br>
{{{abs((3x+4y-12)/5)=(12-3x-4y)/5}}}<br>
and the sum of the distances of P from the three sides of the triangle is<br>
{{{x+y+(12-3x-4y)/5=(12+2x+y)/5}}}<br>
Given that expression for the sum of the distances of P from the three sides of the triangle, it is clear that the minimum sum is at (0,0), where the sum of the distances is 12/5.<br>
For the maximum sum, note that for a given value of x the sum of the distances is greatest is when y is as large as possible; and for a given value of y the sum of the distances is greatest is when x is as large as possible.  That means the maximum sum is when the point P is somewhere on the boundary line 3x+4y=12.<br>
{{{3x+4y=12}}}
{{{4y=-3x+12}}}
{{{y=(-3/4)x+3}}}<br>
The sum of the distance is then<br>
{{{(12+2x+((-3/4)x+3))/5=((5/4)x+15)/5=(1/4)x+3}}}<br>
And clearly the maximum value of that expression is when x is as large as possible -- at (4,0).<br>
The maximum sum is then<br>
{{{(12+2x+y)/5=(12+8)/5=20/5=4}}}<br>
ANSWERS:
minimum sum 12/5 = 2.4, at (0,0)
maximum sum 4, at (4,0)<br>