Posted by **Riley** on Wednesday, April 22, 2009 at 10:44am.

Find an equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

How do I solve? I am stuck. The only thing I can come with is this:

y-5=m(x-3)

How to use this I don't know.

- Calculus -
**Reiny**, Wednesday, April 22, 2009 at 11:37am
draw a line through (3,5) cutting the positive x and y axes at (a,0) and (0,b)

then the area of the triangle is

A = ab/2

but the slope of the two segments must be equal, so 5/(3-a) = (b-5)/-3

solving for b gave me b = 5a/(a-3)

then A = 5a^2/(2a-6)

using the quotient rule

dA/da = [(2a-6)(10a) - 5a^2(2)]/(2a-6)^2

= 0 for a max/min of A

simplifying the top and setting it equal to zero gave me

a^2 - 6a = 0

a(a-6)=0

a = 0 or a = 6

clearly a=0 does not give me a triangle, so a = 6 and b= 10

so slope = 5/-3

and your equation would be

y-5 = (-5/3)(x-3)

take it from there.

- Calculus -
**Riley**, Wednesday, April 22, 2009 at 11:46am
how did you get the slope of the two lines? I get everything else though.

- Calculus -
**Reiny**, Wednesday, April 22, 2009 at 12:02pm
there is only one line

you were given the point (3,5) and I found

the x-intercept to be (6,0)

so slope = (5-0)/(3-6)

= 5/-3 = - 5/3

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