What is the maximum vertical distance between the line
y = x + 72
and the parabola
y = x2 for −8 ≤ x ≤ 9?
what is the maximum vertical distance between the line
y = x + 72
and the parabola
y = x^2 for −8 ≤ x ≤ 9?

the maximum distance is and can be found at x = 

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

Therefore:

Hence, the maximum distance is  = 72.25 and can be found at x = 

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

Since the second derivative:

d"(x) = -2 

is negative, the point is a maximum.

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

Hence, the maximum distance is  = 72.25 and can be found at x = 

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared

a

step-by-step explanation: