Question

Find the equation of the line through the point (3, 5) that cuts off the least area from the first quadrant?
y-A=B(x-C)
i know its slope is -3/5 which is B. But what are the values of A and C?

Answer 1

How did you get the slope to be -3/5 ??

let the x-intercept be A(a,0) and the y-intercept be B(0,b)
label point (3,5) as P

By ratios in similar triangles:
(a-3)/5 = 3/(b-5)
ab-5a-3b+5 = 15
b(a-3) = 5a
b = 5a/(a-3)

Area of triangle = (1/2)ab
= (1/2)(a)(5a/(a-3) )
= 5a^2/(2a-6)
d(Area)/da = ( (2a-6)(10a) - 5a^2 (2))/(2a-6)^2
= (20a^2 - 60a - 10a^2)/(2a-6)^2
= 0 for a min of area

10a^2 - 60a = 0
a(10a - 60) = 0
a = 0 or a = 6
if a = 6, the point A is (6,0) and the slope of the line is
(0-5)/(6-3) = -5/3

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

y = (-5/3)x + 10

check: according to my equation the y-intercept should be B(0,10)
from my b = 5a/(a-3)
b = 30/(3) = 10
looks good

Answer 2

Yes i did exactly the same but website is not accepting my answer, not sure why

Answer 3

You did not do exactly the same, your slope was -3/5, different from mine.

(another example of why I prefer traditional class room instructions, where the learning of math is not just "answer based" )