
New Member
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
[tex]y=96x[/tex]
Distance Formula:
[tex]d=sqrt{(x+3)^2 + (y1)^2}[/tex]
[tex]d(x)=sqrt{(x+3)^2 + (96x1)^2}[/tex]
[tex]d(x)=sqrt{37x^290x+73}[/tex]
[tex]f(x)=d(x)^2=37x^290x+73[/tex]
derivitive:
[tex]f'(x)=74x90[/tex]
Set the derivative equal to 0:
[tex]0=74x90[/tex]
[tex]x=45/37[/tex]
That is how far I can get by looking at the examples in the book. Then I get kinda lost... Any ideas? Thank you!

Senior Member
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

New Member

Elite Member
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
[tex]\L\\S=L^{2}=(x+3)^{2}+(\underbrace{(96x)}_{\text{y}}1)^{2}[/tex]
[tex]\L\\S=37x^{2}90x+73[/tex]
[tex]\L\\S'=7490x[/tex]
[tex]\L\\x=\frac{45}{37}, \;\ y=\frac{63}{37}[/tex]
Excuse me, skeet, but I think the yintercept of that line should be 3/2, not 1/2.
Hence the discrepancy. [tex]y=\frac{1}{6}x+\frac{3}{2}[/tex]
gives the same result as the calc way.

New Member

New Member
ok im stuck again
Find the dimensions of a rectangle with area 1000m^2 whose perimeter is small as possible.
[tex]Area=x*y[/tex]
[tex]1000=x*y[/tex]
[tex]y=1000x[/tex]
[tex]Permiter=2x*2y[/tex]
im stuck on how these releate to each other
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