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lidiya [134]
3 years ago
5

Quick Question!! Help me out :)

Mathematics
1 answer:
Kipish [7]3 years ago
4 0

Answer:

R:  (16, 2)

Step-by-step explanation:

Let (x, y) be the point R

(x - 10)/2 = 3       and (y + 12)/2 = 7

x - 10 = 6                     y + 12 = 14

x = 16                           y = 2

R:  (16, 2)

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Answer:

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Step-by-step explanation:

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Line segment 19 units long running from (x,0) ti (0, y) show the area of the triangle enclosed by the segment is largest when x=
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The area of the triangle is

A = (xy)/2

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y = sqrt(361 - x^2)

Now we substitute this expression for y in the area equation.

A = (1/2)(x)(sqrt(361 - x^2))

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We take the derivative of A with respect to x.

dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]

dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]

Now we set the derivative equal to zero.

(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0

-2x^2 + 361 = 0

-2x^2 = -361

2x^2 = 361

x^2 = 361/2

x = 19/sqrt(2)

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(19/sqrt(2))^2 + y^2 = 361

361/2 + y^2 = 361

y^2 = 361/2

y = 19/sqrt(2)

We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
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3 years ago
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