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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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)

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)

Carpanedam7

Answered 2022-08-02
Author has **3** answers

The distance formula is as follows:

$d=\sqrt{(\mathrm{\u25b3}x{)}^{2}+(\mathrm{\u25b3}y{)}^{2}}=\sqrt{({x}_{2}-{x}_{1}{)}^{2}+({y}_{2}-{y}_{1}{)}^{2}}$

Using this formula, we can calculate the distances between thepoints.

The distance between point A&B is:

$\sqrt{(x-0{)}^{2}+(0-3{)}^{2}}=\sqrt{{x}^{2}+9}$

Between B&C:

$\sqrt{(20-x{)}^{2}}=20-x$

So we know that $90,000(\sqrt{{x}^{2}+9})+48,000(20-x)=1,500,000$

$90,00(\sqrt{{x}^{2}+9})+960,000-48,000x=1,500,000$

$90,000(\sqrt{{x}^{2}+9})=540,000+48,000x$

$\sqrt{{x}^{2}+9}=6+\frac{8}{15}x$

${x}^{2}+9=\frac{64}{225}{x}^{2}+\frac{32}{5}x+36$

$0=-\frac{161}{225}{x}^{2}+\frac{32}{5}x+36$

$0=-\frac{161}{225}{x}^{2}+\frac{32}{5}x+27$

Now if we use the quadratic equation

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$, we get

$\frac{-\frac{32}{5}\pm \sqrt{(\frac{32}{5}{)}^{2}-4(\frac{-161}{225})(27)}}{2(\frac{-161}{225})}$

$\frac{720}{161}\pm (-\frac{225}{322}\sqrt{\frac{2956}{25}})$

$\therefore $ b(x)=-3.13 (-3.13,0)

or b(x)=12.1 (12.1,0)

$d=\sqrt{(\mathrm{\u25b3}x{)}^{2}+(\mathrm{\u25b3}y{)}^{2}}=\sqrt{({x}_{2}-{x}_{1}{)}^{2}+({y}_{2}-{y}_{1}{)}^{2}}$

Using this formula, we can calculate the distances between thepoints.

The distance between point A&B is:

$\sqrt{(x-0{)}^{2}+(0-3{)}^{2}}=\sqrt{{x}^{2}+9}$

Between B&C:

$\sqrt{(20-x{)}^{2}}=20-x$

So we know that $90,000(\sqrt{{x}^{2}+9})+48,000(20-x)=1,500,000$

$90,00(\sqrt{{x}^{2}+9})+960,000-48,000x=1,500,000$

$90,000(\sqrt{{x}^{2}+9})=540,000+48,000x$

$\sqrt{{x}^{2}+9}=6+\frac{8}{15}x$

${x}^{2}+9=\frac{64}{225}{x}^{2}+\frac{32}{5}x+36$

$0=-\frac{161}{225}{x}^{2}+\frac{32}{5}x+36$

$0=-\frac{161}{225}{x}^{2}+\frac{32}{5}x+27$

Now if we use the quadratic equation

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$, we get

$\frac{-\frac{32}{5}\pm \sqrt{(\frac{32}{5}{)}^{2}-4(\frac{-161}{225})(27)}}{2(\frac{-161}{225})}$

$\frac{720}{161}\pm (-\frac{225}{322}\sqrt{\frac{2956}{25}})$

$\therefore $ b(x)=-3.13 (-3.13,0)

or b(x)=12.1 (12.1,0)

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