Asked by Emily
Find the slope and y-intercept of the line through the point (6,2) that cuts off the least area from the first quadrant.
Answers
Answered by
Damon
y = m x + b
2 = 6 m + b
m=(2-b)/6
when y = 0
x = -b/m
so
Area = (1/2)b(-b/m)
A = -b^2/2m
A = -3b^2/(2-b)
dA/db = [(2-b)(-6b)+3b^2(-1)]/bot^2
=0 for min or max
3 b^2 = -12b + 6 b^2
3 b^2 = 12 b
b = 4 (or zero of course)
so y intercept is 4
m = (2-b)/6
= -2/6 = -1/3
2 = 6 m + b
m=(2-b)/6
when y = 0
x = -b/m
so
Area = (1/2)b(-b/m)
A = -b^2/2m
A = -3b^2/(2-b)
dA/db = [(2-b)(-6b)+3b^2(-1)]/bot^2
=0 for min or max
3 b^2 = -12b + 6 b^2
3 b^2 = 12 b
b = 4 (or zero of course)
so y intercept is 4
m = (2-b)/6
= -2/6 = -1/3
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