Minimize distance calculus
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Minimize distance calculus

[From: ] [author: ] [Date: 12-07-16] [Hit: ]
but they do not. You need to use the distance formuladist^2 = (y2 - y1)^2 + (x2 - x1)^2So find the minimum of when the distance between (y2 - 3)^2, and (x2 + 5)^2 in the distance formula.-Alternatively you can solve this without calculus since the point on a line closest to an external point always forms a perpendicular line with that point.Simply find a line perpendicular to 4x + y = 6 that passes through (-5, 3).......
D(x) = √( (x - (-5) )^2 + (6 - 4x - 3)^2 )
D(x) = √( (x + 5)^2 + (-4x + 3)^2 )
D(x) = √( x^2 + 10x + 25 + 16x^2 - 24x + 9 )
D(x) = √( 17x^2 - 14x + 34 )

Then, find the absolute minimum by setting the derivative equal to zero:
D'(x) = 1/2 * (34x - 14) / √( 17x^2 - 14x + 34 )

0 = 1/2 * (34x - 14) / √( 17x^2 - 14x + 34 )
And everything clears away except:
0 = 34x - 14
solve for x:
34x = 14
x = 7/17

plug 7/17 into the equation of the line (y = -4x + 6) to find the y-coordinate:
y = -4 * 7/17 + 6
y = -28/17 + 102/17
y = 74/17

So the answer is (7/17, 74/17)
To 2 decimal places: (0.41, 4.35)

-
Think about this one though. The second derivitive is actual 0, meaning there is no maximum or minimum.

f(x) = -4x + 6

f'(x) = -4 meaning it is always decreasing

f''(x) = 0 so it will have no change in concavity.

You are assuming they have an intersection, but they do not. You need to use the distance formula

dist^2 = (y2 - y1)^2 + (x2 - x1)^2

So find the minimum of when the distance between (y2 - 3)^2, and (x2 + 5)^2 in the distance formula.

-
Alternatively you can solve this without calculus since the point on a line closest to an external point always forms a perpendicular line with that point.

Simply find a line perpendicular to 4x + y = 6 that passes through (-5, 3).
Then set both lines equal to each other to find the intersection point.

4x + y = 6
y = -4x + 6

Line perpendicular has negative reciprocal slope.
y = 1/4x + b

Plug in (-5, 3) to get b:

3 = 1/4(-5) + b

12/4 = -5/4 + b

17/4 = b

y = 1/4x + 17/4

Now set both lines equal:


-4x + 6 = 1/4x + 17/4
-16/4x + 24/4 = 1/4x + 17/4
7/4 = 17/4*x
7 = 17x

x = 7/17

Now plug into line equation to get y:
y = -4x + 6
y = -4(7/17) + 6
y = -28/17 + 102/17
y = 74/17

Point closest:

(7/17, 74/17)
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keywords: distance,Minimize,calculus,Minimize distance calculus
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