Question

What is the maximum vertical distance between the line

Answer 1

Answer: the maximum distance is $\frac{289}{4}$ and can be found at x = $\frac{1}{2}$

Let's call:

f(x) = x + 72

g(x) = x²

A point belonging to the line will be L(x, x+72) and a point on the parabola will be P(x, x²). It can be easily seen that in the interval -8 ≤ x ≤ 9 the line is above the parabola (it's enough to graph them or plug in some numbers), therefore their distance at any point will be:

d(x) = f(x) - g(x) = - x² + x + 72

The function d(x) is a parabola that opens downward, therefore the maximum will be the vertex; given a parabola

y(x) = ax² + bx + c

the coordinates of the vertex will be

$V(\frac{-b}{2a}, y(\frac{-b}{2a}))$

Therefore:

$V(\frac{1}{2}, \frac{289}{4})$

Hence, the maximum distance is $\frac{289}{4}$ = 72.25 and can be found at x = $\frac{1}{2}$

Another way to find the maximum is to use calculus to find the first derivative of the distance:

d'(x) = -2x + 1

and set it equal to zero:

-2x + 1 = 0

$x = \frac{1}{2}$

Since the second derivative:

d"(x) = -2 

is negative, the point is a maximum.

Then, substitute this value in the equation for the distance:

$d(\frac{1}{2}) = -(\frac{1}{2})^{2} + \frac{1}{2} + 72$

$d(\frac{1}{2}) = \frac{289}{4} = 72.25$

Hence, the maximum distance is $\frac{289}{4}$ = 72.25 and can be found at x = $\frac{1}{2}$