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.

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(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.

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3