Solving for x with ratios and shapes

Question

Q: Consider an experiment in which a marble is tossed into a box whose base is shown in the figure. The probability that the marble will come to rest in the shaded portion of the box is equal to the ratio of the shaded area to the total area of the figure. If the probability is equal to 3/10, find the positive value of x. (If you could show the steps as well, that would be super helpful. I am a bit confused about how to solve this problem.)

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The rectangle: | ------------------------------- |

| | |. |

| | | x-1 |. 2x+1

| | |. |

| ------------------------------- |

| x+2 |

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3x+2

Answer 1

Assuring the inner rectangle is shaded:

(x + 2)(x - 1) / (3x + 2)(2x + 1) = 3 / 10

(x² + x - 2) / (6x² + 7x + 2) = 3 / 10

10x² + 10x - 20 = 18x² + 21x + 6

0 = 8x² + 11x - 26

Discriminant < 0, solutions are complex

Assuming the boarder is shaded

(x + 2)(x - 1) / (3x + 2)(2x + 1) = 10 / 3

Discriminant < 0, solutions are complex

Check the data.

Answer 2

For x positive, it is easy to show that the area of the inner box will be less than 3/10 of the area of the outer box for all x. Since this means that the area outside the inner box will be more than 7/10 of the total area, there will be no solution.

As x is positive, x > 1 for x - 1 > 0, a necessary condition for the inner box to have positive area (the 3 other sides will always be positive).

To show th e 3/10 , we will show that (x-1)/(2x+1) < 1/2 for all positive x, and therefore, for x > 1, and (x+2)/(3x+2) < 3/5 for all x > 1

The product of these fractions is the area of the inner box / the area of the outer box.

As 3/5 * 1/2 = 3/10, this means that the area of the inner box is less than 3/10 of the area of the outer box.

First, we show that (x - 1)/(2x + 1) < 1/2 for all positive x.

(x - 1)/(2x + 1) = (x + 1/2 - 1/2 - 1)/(2x + 1) = (x + 1/2)/(2 x + 1) - (3/2)/(2x + 1) = 1/2 - 3/(4 x + 2)

For x > 0, 4x + 2 > 0, 3/(4x+2) > 0 and 1/2 - 3/(4x+2) < 1/2

(x + 2)/(3x + 2) = (x + 2/3 - 2/3 + 2)/(3 x + 2) = (x + 2/3)/(3x +2) + 4/(9x + 6) = 1/3 + 4/(9x + 6)

For x > 1, 9x + 6 > 0 and increasing, so 4/(9x + 6) is decreasing. Thus, for x > 1, 4/(9x + 6) < 4/(9*1 + 6) = 4/15, and (x + 2)/(3x + 2) = 1/3 + 4/(9x + 6) < 1/3 + 4/15 = 3/5

(x + 2)/(3x + 2) < 3/5 for all x > 1

Thus, we have shown (x + 2)/(3x + 2) < 3/5 for all x > 1, (x - 1)/(2x + 1) < 1/2 for all x > 1, and

(x + 2)(x - 1)/((3x+2)(2x+1)) = area of inner box/area of outer box < 3/10 for all x > 1. As we already showed it is <= 0 for 0 <= x <= 1, area of inner box/area of outer box < 3/10 for all positive x.