Minimum Problem (Calculus) (1 Viewer)

[atar90plus]

4. Two depots A and B are 2km and 1km from highway OM. A supply station P is to be built on the highway such that the sum of the lengths of roads PA and PB is to be least. If OM is 3 km

a) Express AP + PB in terms of x (Hint : Draw a triangle)

b) Find the position of P that gives the least roadway from A to B

 

[Internet_Police]

Draw a straight horizontal line, call it OM and give it a length of 3km. Then go in "x" metres from the left hand side (the point called O) and put the point P

This makes OP= xkm and PM = (3-x)km

Now, I drew points and A and B 2km and 1km vertically above the line OM on different sides of P. Then if you draw the lines from A to P and from B to P you will have two right angled triangles (AOP & BMP).


Now using a^2=b^2 +c^2 , you will get PA = sqrt( x^2+2^2) = sqrt(x^2+4) and PB = sqrt( (3-x)^2 +1^2) = sqrt(x^2 -6x+10)


Therefore, call the distance "d".

d= sqrt(x^2 +4) + sqrt(x^2 -6x+10) = (x^2+4)^(1/2) + (x^2 -6x+10)^(1/2)

Now it's just tedious derivatives and checking that it gives a minimum.