Answer
a. $\sqrt {x^2+6^2}+\sqrt {(12-x)^2+3^2}\ mi$
b. $x=8\ mi$
Work Step by Step
a. Use the figure given in the Exercise and use the Pythagorean Theorem, we have the road length to A is $A=\sqrt {x^2+6^2}$, the road length to B is $B=\sqrt {(12-x)^2+3^2}$, and the total length is $L=A+B=\sqrt {x^2+6^2}+\sqrt {(12-x)^2+3^2}\ mi$
b. Given $L=15\ mi$, we have $\sqrt {x^2+6^2}+\sqrt {(12-x)^2+3^2}=15$ or $\sqrt {(12-x)^2+3^2}=15-\sqrt {x^2+6^2}$. Take the square on both sides to get $(12-x)^2+9=225+x^2+36-30\sqrt {x^2+6^2}$ which is equivalent to $5\sqrt {x^2+6^2}=4x+18$. Take square again to get $9x^2-144x+576=0$ or $(3x-24)^2=0$. Thus $x=8\ mi$ which is at the $\frac{2}{3}$ point of the expressway between A and B.
