Arain gutter is to be made of aluminum sheets that are 12 inches wide by turning up the edges 90degrees. see the illustration. (a) what depth will provide maximum cross-sectional area and hence allow the most water to flow? (b) what depths will allow at least 16 square inches of water to flow?

Step-by-step explanation:

(a) For a depth of x, the two sides of the rain gutter are length x, and the bottom is length (12-2x). The cross sectional area is the product of these dimensions:

  A = x(12 -2x)

This equation describes a parabola that opens downward. It has zeros at ...

  x = 0

  12 -2x = 0 . . . . x = 6

The maximum area is halfway between these zeros, at x=3.

The maximum area is obtained when the depth is 3 inches.

__

(b) For an area of at least 16 square inches, we want ...

  x(12 -2x) ≥ 16

  x(6 -x) ≥ 8 . . . . . divide by 2

  0 ≥ x² -6x +8 . . . . subtract the left side

  (x -4)(x -2) ≤ 0 . . . factor

The expression on the left will be negative for values of x between 2 and 4 (making only the x-4 factor be negative). Hence the the depths of interest are in that range.

At least 16 square inches of water will flow for depths between 2 and 4 inches, inclusive.

c. (3,2,6) is the answer.

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."