Dan O. answered • 12/13/20
Perfect score on Math SAT / GRE; 5 yrs experience as a AP Calc teacher
This is an optimization problem. We want to optimize "distance" from the Distance Formula. Specifically, we want to minimize the distance between (2, -1) and (x,y), where (x,y) is a point on the function.
First, let's rearrange 3x + 2y - 6 = 0 and solve for y.
y = -1.5x + 3
Distance Formula:
D = Square root [(x2 - x1)2 + (y2 - y1)2]
Plugging in our points...
D = Square root [(x - 2)2 + (y - (-1))2]
D = Square root [(x - 2)2 + (y + 1))2]
D = Square root [(x - 2)2 + ((-1.5x + 3) + 1))2] (make a substitution)
D = Square root [(x - 2)2 + (-1.5x + 4)2]
D = Square root (x2 - 4x + 4 + 2.25x2 - 12x + 16)
D = (3.25x2 - 16x + 20)1/2
Maxima and minima occur at points when the function has a horizontal tangent line. Horizontal tangent lines have slopes of zero, therefore we can find x-values where maxima and minima occur by taking the derivative and setting it equal to zero, because the derivative is equal to the slope of the tangent line.
D' = (.5)(3.25x2 - 16x + 20)-1/2 x (6.5x - 16)
This derivative becomes a fraction since we have an expression raised to a negative exponent (it doesn't allow me to write it as a fraction here).
When we set a fraction = 0, we are only concerned with the numerator equaling 0, since 0 on the denominator makes the fraction undefined. So...
6.5x - 16 = 0
x = 32/13
x = 2.461538462
1st derivative test (number line):
D'(1) = -
D'(2.461538) = 0
D'(3) = +
Because D'(x) switches from - to + @ x = 32/13, D(x) switches from decreasing to increasing @ x = 32/13, therefore a minimum value on D(x) occurs at x = 32/13.
D(32/13) = -1.5(32/13) + 3
= -48/13 + 39/13
= -9/13
The point on the line "3x + 2y - 6 = 0" which is closest to (2, -1) is the point (32/13, -9/13).
