# Thread: Find pt. on 6x + y = 9 closest to (-3, 1) in terms of y

1. ## Find pt. on 6x + y = 9 closest to (-3, 1) in terms of y

Find the point on the line 6x + y = 9 closest to the point (-3, 1) in terms of y

$y=9-6x$

Distance Formula:

$d=sqrt{(x+3)^2 + (y-1)^2}$

$d(x)=sqrt{(x+3)^2 + (9-6x-1)^2}$

$d(x)=sqrt{37x^2-90x+73}$

$f(x)=d(x)^2=37x^2-90x+73$

derivitive:

$f&#39;(x)=74x-90$

Set the derivative equal to 0:

$0=74x-90$

$x=45/37$

That is how far I can get by looking at the examples in the book. Then I get kinda lost... Any ideas? Thank you!

2. 6x + y = 9

slope of this line is m = -6

a perpendicular line has slope m = 1/6

the line passing through (-3,1) with slope m = 1/6 is

y - 1 = (1/6)(x + 3) edit mistake ... I wrote y + 1 instead of y - 1 on the left side of the linear equation
y = (1/6)x + (3/2)

the point of intersection of the line y = (1/6)x + (3/2) and 6x + y = 9 is ...

9 - 6x = (1/6)x + (3/2)
54 - 36x = x + 9
45 = 37x

x = 45/37, y = 9 - 6(45/37) = 63/37 now we should all agree

3. that was simple

thanks

4. I get the same thing as gopher.

When you are maximizing or minimizing a distance, there is a trick you can use to avoid radicals. The distance and the square of the distance have their max and min at the same point. Therefore, the minimum occurs at

$\L\\S=L^{2}=(x+3)^{2}+(\underbrace{(9-6x)}_{\text{y}}-1)^{2}$

$\L\\S=37x^{2}-90x+73$

$\L\\S&#39;=74-90x$

$\L\\x=\frac{45}{37}, \;\ y=\frac{63}{37}$

Excuse me, skeet, but I think the y-intercept of that line should be 3/2, not 1/2.
Hence the discrepancy. $y=\frac{1}{6}x+\frac{3}{2}$
gives the same result as the calc way.

5. very good thank you both

6. ok im stuck again
Find the dimensions of a rectangle with area 1000m^2 whose perimeter is small as possible.

$Area=x*y$
$1000=x*y$
$y=1000x$

$Permiter=2x*2y$

im stuck on how these releate to each other

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