# Installing an electric power line underwater costs $90,000 permile and running it along land costs$48,000 per mile. Supposepoint A is at coordinates (in miles) (0,3), point B is at (x,0),and point C is at coordinates (20,0). To run the wire from point Ato point B, the power line must be run underwater. To run the powerline from point B to point C, the power line runs along land. Thetotal cost to run the power line from point A to point B to point C is $1,500,000. Where is point B? Freddy Friedman 2022-07-31 Answered Installing an electric power line underwater costs$90,000 permile and running it along land costs $48,000 per mile. Supposepoint A is at coordinates (in miles) (0,3), point B is at (x,0),and point C is at coordinates (20,0). To run the wire from point Ato point B, the power line must be run underwater. To run the powerline from point B to point C, the power line runs along land. Thetotal cost to run the power line from point A to point B to point C is$1,500,000. Where is point B?
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## Answers (2)

Helena Howard
Answered 2022-08-01 Author has 12 answers
$1,500,000 = (x/y)*$90,000 + a*$48,000 because A and B have a distance in the x and y direction betweenthem, while B and C only have distance in the x direction betweenthem, a distance that could be either (x-20) or (20-x) depending onwhere its located (we are going to call this a for now. now we solve for x assuming a=(x-20)..... so x is greater than20$1,500,000 = (x/y)*$90,000 + (x-20)*$48,000
$1,500,000 = x*((1/y)*$90,000 + $48,000) -(20)$48,000.................... distribute the $48000 and take outthe x form the rest of the terms. x = ($1,500,000+(20)$48000) / ((1/y)*$90,000 + $48,000) now plug in 3 for y and you get x = 31.54 or round it to 32 now we try it with a = (20-x)$1,500,000 = (x/y)*$90,000 + (20-x)*$48,000
$1,500,000 = x*((1/y)*$90,000 - $48,000) +(20)$48,000.................... distribute the $48000 and take outthe x form the rest of the terms. x = ($1,500,000-(20)$48000) / ((1/y)*$90,000 + $48,000) now plug in 3 for y and you get x = -30 so you get two answers B = (32,0) and (-30,0) so i guess your answer will depend on whether B or C are fartherform A, but they both work.... you can try to plug the answers in and see that you get$1,500,000in both (just remember that the equation depends on where you wantpoint B to be with relation to point C)
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Carpanedam7
Answered 2022-08-02 Author has 3 answers
The distance formula is as follows:
$d=\sqrt{\left(\mathrm{△}x{\right)}^{2}+\left(\mathrm{△}y{\right)}^{2}}=\sqrt{\left({x}_{2}-{x}_{1}{\right)}^{2}+\left({y}_{2}-{y}_{1}{\right)}^{2}}$
Using this formula, we can calculate the distances between thepoints.
The distance between point A&B is:
$\sqrt{\left(x-0{\right)}^{2}+\left(0-3{\right)}^{2}}=\sqrt{{x}^{2}+9}$
Between B&C:
$\sqrt{\left(20-x{\right)}^{2}}=20-x$
So we know that $90,000\left(\sqrt{{x}^{2}+9}\right)+48,000\left(20-x\right)=1,500,000$
$90,00\left(\sqrt{{x}^{2}+9}\right)+960,000-48,000x=1,500,000$
$90,000\left(\sqrt{{x}^{2}+9}\right)=540,000+48,000x$
$\sqrt{{x}^{2}+9}=6+\frac{8}{15}x$
${x}^{2}+9=\frac{64}{225}{x}^{2}+\frac{32}{5}x+36$
$0=-\frac{161}{225}{x}^{2}+\frac{32}{5}x+36$
$0=-\frac{161}{225}{x}^{2}+\frac{32}{5}x+27$
Now if we use the quadratic equation
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$, we get
$\frac{-\frac{32}{5}±\sqrt{\left(\frac{32}{5}{\right)}^{2}-4\left(\frac{-161}{225}\right)\left(27\right)}}{2\left(\frac{-161}{225}\right)}$
$\frac{720}{161}±\left(-\frac{225}{322}\sqrt{\frac{2956}{25}}\right)$
$\therefore$ b(x)=-3.13 (-3.13,0)
or b(x)=12.1 (12.1,0)
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