## anonymous one year ago Two towns, A and B, are 13.0 miles apart and located 8.0 and 3.0 miles east, respectively, of a long, straight highway. A construction company has a contract to build a road from town A to the highway and then to town B. Determine the length (to the nearest tenth of a mile) of the shortest road that meets these requirements.

• This Question is Open
1. anonymous

@ganeshie8 can you draw it picture,so that i can solve?

2. ganeshie8

|dw:1436781239171:dw|

3. ganeshie8

|dw:1436782203008:dw|

4. ganeshie8

I think we need to find the minimum length of that road

5. anonymous

did they told us that the road should meet at the same point to highway?

6. anonymous

|dw:1437272681909:dw| now express distance of 2 roads as function of x $f(x) = \sqrt{x^2 +8^2} + \sqrt{(12-x)^2 +3^2}$ minimize function by taking derivative and setting equal to 0 (use chain rule) $f'(x) = \frac{x}{\sqrt{x^2+64}} - \frac{12-x}{\sqrt{(12-x)^2 + 9}} = 0$ Solve for x first square both sides $\frac{x^2}{x^2+64} = \frac{(12-x)^2}{(12-x)^2 +9}$ $x^2 (x^2 -24x +153) = (x^2 +64)(x^2 -24x+144)$ $57x^2 -(64)(24)x +(64)(144) = 0$ $19x^2 - 512x+3072 = 0$ At this point just plug it into quadratic formula (remember x < 12) $x = 9.02$ Now plug this value into f(x) to get length of shortest road