A barn is located on the next concession from the house, as shown in the diagram. A water line must be put in from the house to the barn. If it is run through the field, it costs twice as much...

A barn is located on the next concession from the house, as shown in the diagram. A water line must be put in from the house to the barn. If it is run through the field, it costs twice as much as it does when the line runs along the road. Explain the cost function, c(x)=(2-x)+2√(x^2+(.05)^2) , and analyze it to find the best place to put the line. Analysis the derivative of C(x) with respect to the minimum cost. 

 

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pramodpandey's profile pic

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We have given cost  function as

`C(x)=(2-x)+2sqrt(x^2+(.05)^2)`

differentiate C(x),with respect to x,to get critical points

`C'(x)=-1+2(1/2)1/(sqrt(x^2+(.05)^2)) (2x)`

`C'(x)=0`    if

`sqrt(x^2+(.05)^2)-2x=0`

`x^2+(.05)^2=4x^2`

`x=sqrt((.05)^2/3)=.029`

To check is x=.029 is point of minima,find second derivative of C(x),

`C''(xx)=2(sqrt(x^2+(.05)^2)-x(1/2)(1/sqrt(x^2+(.05)^2))(2x))/(x^2+(.05)^2)`

`C''(x)={2(x^2+(.05)^2-x^2)}/((x^2+(.05)^2)^(3/2))`

Thus

`C''(x)}_{x=.029}>0`

Thus x=.029 will give minimum cost. A water line must be  at

x=.029

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