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.
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.
how did you get the slope of the two lines? I get everything else though.
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
To find the equation of the line that cuts off the least area in the first quadrant and passes through the point (3,5), you can follow these steps:
Step 1: Assume an initial equation:
Start by assuming the equation of the line is in the standard form, y = mx + b, where m represents the slope of the line, and b represents the y-intercept.
Step 2: Determine the optimal slope:
Since the line cuts off the least area in the first quadrant, it will be perpendicular to the line passing through the origin (0,0) and the given point (3,5). The slope of the line passing through the origin and (3,5) can be calculated as:
m1 = (5 - 0) / (3 - 0) = 5/3.
Since the line we are looking for is perpendicular to this line, the slope will be the negative reciprocal of m1:
m2 = -1 / m1 = -3/5.
Step 3: Substitute the slope and point into the equation:
Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point, substitute the values:
y - 5 = (-3/5)(x - 3).
Step 4: Simplify the equation:
To convert the equation into slope-intercept form (y = mx + b), expand the right side of the equation and simplify:
y - 5 = (-3/5)x + (9/5)
y = (-3/5)x + (9/5) + 5
y = (-3/5)x + (9/5) + 25/5
y = (-3/5)x + (34/5).
Therefore, the equation of the line that passes through the point (3,5) and cuts off the least area in the first quadrant is:
y = (-3/5)x + (34/5).