Exercise 32 Page 936

We are given a system of inequalities. We want to find the probability that if a point

(x, y)

is chosen at random in solution set of the system, it is a solution to the inequality

(x - 1)^{2} + (y - 1)^{2} \geq 16 .

First, let's take a look at the given system.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 1 \leq x \leq 6 y \leq x y \geq 1

Since the first inequality is compound, we can rewrite it as two separate inequalities.

⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ x \leq 6 x \geq 1 y \leq x y \geq 1

Let's graph all the inequalities in one coordinate system.

The solution set of our system is a right triangle. Note that the length of both legs is $5 .$ This means that it is a $4 5^{\circ} - 4 5^{\circ} - 9 0^{\circ}$ triangle.

Now, let's also graph the inequality $(x - 1)^{2} + (y - 1)^{2} \geq 16 .$ The equation of the boundary line of this inequality is an equation of a circle with center $(1, 1)$ and radius $4$ in standard form.

We can see that the intersection of the circle and the triangle is a sector of the circle. To find the probability that the point satisfies the given inequality, we will use geometric probability. The probability that a point chosen at random in the triangle satisfies the given inequality is the ratio of the area

A

of the region of the triangle outside the circle to the area

A_{T}

of the triangle .

P ((x - 1)^{2} + (y - 1)^{2} \geq 16) = \frac{A}{A _{T}}

First, let's find the area of the triangle. We know the formula for the area

A_{T}

of a right triangle with legs of length

a

and

b .

A_{T} = \frac{1}{2} a b

In our case, both legs have a length of

5 .

By substituting

5

for both

a

and

b

in this formula, we can calculate the area

A_{T}

of our triangle.

A_{T} = \frac{1}{2} a b

SubstituteII

$a = 5$ , $b = 5$

A_{T} = \frac{1}{2} (5) (5)

Multiply

A_{T} = 12.5

To find the area

A

of the region of the triangle that is outside of the circle, we will subtract the area

A_{S}

of the sector of our circle from the area

A_{T}

of the triangle. Recall the formula for the area of a sector of a circle with measure

θ

and radius

r .

A_{S} = \frac{θ}{3 6 0} π r^{2}

In our case, the measure of the sector is

4 5^{\circ}

and the radius is

4 .

By substituting these values for

θ

and

r,

respectively, in the formula, we can calculate the area

A_{S}

of the sector.

A_{S} = \frac{θ}{3 6 0} π r^{2}

SubstituteII

$θ = 45$ , $r = 4$

A_{S} = \frac{4 5}{3 6 0} π 4^{2}

CalcPowProd

Calculate power and product

A_{S} = 2 π

UseCalc

Use a calculator

A_{S} = 6.283185 \dots

RoundDec

Round to $1$ decimal place(s)

A_{S} \approx 6.3

Now we can calculate the area

A

of the region of the triangle outside the circle.

A = A_{T} - A_{S}

SubstituteII

$A_{T} = 12.5$ , $A_{S} = 6.3$

A = 12.5 - 6.3

SubTerm

Subtract term

A = 6.2

The area of the region of the triangle outside the circle is

6.2 .

Let's calculate our probability!

P ((x - 1)^{2} + (y - 1)^{2} \geq 16) = \frac{A}{A _{T}}

SubstituteII

$A = 6.2$ , $A_{T} = 12.5$

P ((x - 1)^{2} + (y - 1)^{2} \geq 16) = \frac{6 . 2}{1 2 . 5}

CalcQuot

Calculate quotient

P ((x - 1)^{2} + (y - 1)^{2} \geq 16) = 0.496

WritePercent

Convert to percent

P ((x - 1)^{2} + (y - 1)^{2} \geq 16) = 49.6 %

The probability that a point

(x, y)

chosen randomly in the solution set of the system of inequalities is a solution to the inequality

(x - 1)^{2} + (y - 1)^{2} \geq 16

is approximately

0.496,

or approximately

49.6 % .