Mathematics, 19.12.2020 14:00 autumnguidry6867

Consider points A(1, 6) and B(8, 8). Find point C on the x-axis so AC +BC is a minimum. The point that minimizes AC +BC is C (_ , _)

Answers: 1

Show answers

Answers

Answer from: lazybridplayer

The coordinates of a point is the position of the point, on the coordinate plane.

The point C that minimizes AC + BC are (-20, 0) and (4, 0)

The given parameters are:

$\mathbf{A = (1,6)}$

$\mathbf{B = (8,8)}$

From the question, we understand that point C is on the x-axis.

This means that:

$\mathbf{C = (x,0)}$

Calculate segments AC and BC using the following distance formula

$\mathbf{d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}$

So, we have:

$\mathbf{AC = \sqrt{(1 - x)^2 + (6 - 0)^2}}$

$\mathbf{AC = \sqrt{(1 - x)^2 + 6^2}}$

$\mathbf{BC = \sqrt{(8 - x)^2 + (8 -0)^2}}$

$\mathbf{BC = \sqrt{(8 - x)^2 + 8 ^2}}$

Add both distance

$\mathbf{AB + BC = \sqrt{(1 - x)^2 + 6^2} + \sqrt{(8 - x)^2 + 8^2}}$

Differentiate, to calculate the minimum distance

$\mathbf{d' = (1 - x) \times 2 \times (0.5 - 1) \times ((1 - x)^2 + 6^2)^({0.5 - 1}) + (8 - x) \times 2 \times (0.5 - 1) \times ((8 - x)^2} + 8^2))}^{(0.5 - 1)} }$

Simplify

$\mathbf{d' = -\frac{1 - x}{\sqrt{(1 - x)^2 + 6^2}}- \frac{8 - x}{\sqrt{(8 - x)^2 + 8^2}}}$

Set to 0

$\mathbf{ -\frac{1 - x}{\sqrt{(1 - x)^2 + 6^2}} - \frac{8 - x}{\sqrt{(8 - x)^2 + 8^2}} = 0}$

Rewrite as:

$\mathbf{ -\frac{1 - x}{\sqrt{(1 - x)^2 + 6^2}} = \frac{8 - x}{\sqrt{(8 - x)^2 + 8^2}} }$

Cross multiply

$\mathbf{(8 - x) \times\sqrt{(1 - x)\² + 6\²} = -(1 - x) \times\sqrt{(8 - x)\² + 8\²}}$

Square both sides

$\mathbf{(8 - x)^2 \times (1 - x)\² + 6\² = (-(1 - x))^2 \times (8 - x)\² + 8\²}$

$\mathbf{(8 - x)^2 \times (1 - x)\² + 6\² = (1 - x)^2 \times (8 - x)\² + 8\²}$

Expand

$\mathbf{x^4 - 18x^3 + 133x^2 -720x + 2368 = x^4 - 18x^3 + 161x^2 - 272x + 128}$

Evaluate like terms

$\mathbf{133x^2 -720x + 2368 = 161x^2 - 272x + 128}$

Collect like terms

$\mahbf{161x^2 - 133x^2 - 272x + 720x + 128 - 2368 = 0}$

$\mathbf{28x^2 + 448x - 2240 = 0}$

Factor out 28

$\mathbf{28(x^2 + 16x - 80) = 0}$

Divide through by 28

$\mathbf{x^2 + 16x - 80 = 0}$

Expand

$\mathbf{x^2 + 20x - 4x - 80 = 0}$

Factorize

$\mathbf{x(x + 20) + 4(x + 20) = 0}$

Factor out x + 20

$\mathbf{(x + 20)(x - 4) = 0}$

Solve for x

$\mathbf{x =- 20\ or\ x = 4}$

Hence, the point C that minimizes AC + BC are (-20, 0) and (4, 0)

Another question on Mathematics

Mathematics, 21.06.2019 15:30

Amerchant has coffee worth $60 a pound that she wishes to mix with 50 pounds of coffee worth $90 a pound to get a mixture that she will sell for $70 a pound. how many pounds of the & 60 coffee should be used?

Answers: 3

Answer

Mathematics, 21.06.2019 20:30

Select all expressions that are equivalent to 2(3x + 7y). question 1 options: 6x + 14y 6x + 7y 1(6x + 14y)

Answers: 1

Answer

Mathematics, 21.06.2019 23:00

Events a and b are independent. the probability of a occurring is 2/5. the probability of b occurring is 1/4. what is p(a and b)?

Answers: 2

Answer

Mathematics, 22.06.2019 01:00

1) your friend taylor missed class today and needs some identifying solutions of systems. explain to taylor where to find the solution(s) of any system using its graph 2)taylor graphs the following in desmos and decides that f(x) = g(x) at x = 0, x = 1, & x = 3. f(x) = 2x+1g(x) = 2x2+1 provide taylor some feedback that explains which part of the answer was incorrect and why.

Answers: 1

Answer

You know the right answer?

Consider points A(1, 6) and B(8, 8). Find point C on the x-axis so AC +BC is a minimum. The point t...

Questions