Applied Maximum and Minimum Problems

Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=10 - x^2. What are the dimensions of such a rectangle with the greatest possible area?

Width = _____

Height = _____

Answer 1

parabola y=10-x^2

an inscribed rectangle with base on the x axis

what is the maximum volume and dimensions?

vertex or maximum point of the parabola is the point (0,10), axis of symmetry is x=0

maximize the area between y=0, x=0 and the parabola, then double it to get max area

area of that Quadrant I area = xy = x(10-x^2) = 10x-x^3

take the derivative and set it equal to zero

y' = 10-3x^2 = 0

3x^2 = 10

x^2 = 10/3

x = sqr(10/3) = about 1.83 = half width, full width= 2.66

y = 10-1.83^2= 10-10/3 = 20/3 = about 6.67 Length

half the rectangle's area = xy = 1.83(20/3) = about 12.17

double it = 24.34 = max area of the inscribed rectangle

it may help to graph the parabola and rectangle

x intercepts of the parabola are found by sort of factoring

y = 10-x^2 = (sqr10-x)(sqr10+x)

zeros or x intercepts are sqr10 = about 3.16

minimum area = nearly zero with dimensions about 0 by 10 or 6.32 by 0

max area very roughly is the average or 3.16 by 5 high. 2.7 by 6.7 is very roughly in that range