Question

(20 points)
centerville is the headquarters of greedy cablevision inc. the cable company is about to expand service to two nearby towns,
springfield and shelbyville. there needs to be cable connecting centerville to both towns. the idea is to save on the cost of
cable by arranging the cable in a y - shaped configuration. centerville is located at (12, 0) in the xy - plane, springfield is at (0, 2),
and shelbyville is at (0, - 2). the cable runs from centerville to some point (x, 0) on the x - axis where it splits into two branches
going to springfield and shelbyville. find the location (x, 0) that will minimize the amount of cable between the 3 towns and
compute the amount of cable needed. justify your answer.
to solve this problem we need to minimize the following function of x:
f(x)=□
we find that f(x) has a critical number at x = □
to verify that f(x) has a minimum at this critical number we compute the second derivative f(x) and find that its value at the
critical number is □, a positive number.
thus the minimum length of cable needed is □

Explanation:

Step1: Find the function \( f(x) \)

The distance from Centerville \((12,0)\) to \((x,0)\) is \( d_1=|12 - x|=12 - x\) (since \(x<12\)).
The distance from \((x,0)\) to Springfield \((0,2)\) is \(d_2=\sqrt{(x - 0)^2+(0 - 2)^2}=\sqrt{x^{2}+4}\).
The distance from \((x,0)\) to Shelbyville \((0,-2)\) is \(d_3=\sqrt{(x - 0)^2+(0+2)^2}=\sqrt{x^{2}+4}\).
So \(f(x)=(12 - x)+2\sqrt{x^{2}+4}\).

Step2: Find the first - derivative \(f^{\prime}(x)\)

Using the sum rule and the chain rule.
The derivative of \(y = 12 - x\) is \(y^{\prime}=-1\).
The derivative of \(y = 2\sqrt{x^{2}+4}=2(x^{2}+4)^{\frac{1}{2}}\). By the chain rule \(y^{\prime}=2\times\frac{1}{2}(x^{2}+4)^{-\frac{1}{2}}\times2x=\frac{2x}{\sqrt{x^{2}+4}}\).
So \(f^{\prime}(x)=-1+\frac{2x}{\sqrt{x^{2}+4}}\).

Step3: Find the critical number

Set \(f^{\prime}(x) = 0\), then \(-1+\frac{2x}{\sqrt{x^{2}+4}}=0\).
\(\frac{2x}{\sqrt{x^{2}+4}}=1\), \(2x=\sqrt{x^{2}+4}\).
Squaring both sides \(4x^{2}=x^{2}+4\), \(3x^{2}=4\), \(x^{2}=\frac{4}{3}\), \(x=\frac{2}{\sqrt{3}}\approx1.15\) (we take the positive value since \(x>0\) in the context of the problem).

Step4: Find the second - derivative \(f^{\prime\prime}(x)\)

Using the quotient rule. If \(y=\frac{2x}{\sqrt{x^{2}+4}}=(2x)(x^{2}+4)^{-\frac{1}{2}}\).
By the product rule \(y^{\prime}=2(x^{2}+4)^{-\frac{1}{2}}+(2x)\times(-\frac{1}{2})(x^{2}+4)^{-\frac{3}{2}}\times2x\).
\(y^{\prime}=\frac{2}{\sqrt{x^{2}+4}}-\frac{2x^{2}}{(x^{2}+4)^{\frac{3}{2}}}=\frac{2(x^{2}+4)-2x^{2}}{(x^{2}+4)^{\frac{3}{2}}}=\frac{8}{(x^{2}+4)^{\frac{3}{2}}}\).
When \(x = \frac{2}{\sqrt{3}}\), \(f^{\prime\prime}(\frac{2}{\sqrt{3}})=\frac{8}{((\frac{4}{3})+4)^{\frac{3}{2}}}=\frac{8}{(\frac{16}{3})^{\frac{3}{2}}}=\frac{8}{\frac{64}{3\sqrt{3}}}=\frac{3\sqrt{3}}{8}>0\).

Step5: Find the minimum length of cable

Substitute \(x=\frac{2}{\sqrt{3}}\) into \(f(x)\).
\(f(\frac{2}{\sqrt{3}})=12-\frac{2}{\sqrt{3}}+2\sqrt{(\frac{2}{\sqrt{3}})^{2}+4}\).
\(=12-\frac{2}{\sqrt{3}}+2\sqrt{\frac{4 + 12}{3}}=12-\frac{2}{\sqrt{3}}+2\times\frac{4}{\sqrt{3}}=12+\frac{6}{\sqrt{3}}=12 + 2\sqrt{3}\approx12 + 3.46=15.46\).

Answer:

\(f(x)=(12 - x)+2\sqrt{x^{2}+4}\); \(x=\frac{2}{\sqrt{3}}\); \(f^{\prime\prime}(x)=\frac{8}{(x^{2}+4)^{\frac{3}{2}}}\), and at \(x = \frac{2}{\sqrt{3}}\), \(f^{\prime\prime}(\frac{2}{\sqrt{3}})>0\); the minimum length of cable is \(12 + 2\sqrt{3}\).