1. ## An inequalalities/area problem

Agent scully has discovered mysterious markings in an area bounded by (the ">" is underlined) y>0, (the ">" is underlined) 2x+5y>10, and (the ">" is underlined) x+3y<9. She need to mark off the exact area (to the nearest hundreth of a unit) with fencing. How many units of fencing will she need, to the nearest hundreth? Agent Mulder, thinking he can blanket the entire region, needs to know the area of the region,to the nearest hundreth of a unit. What is the total area of covering he will need?

2. Originally Posted by chibix3
Agent scully has discovered mysterious markings in an area bounded by (the ">" is underlined) y>0, (the ">" is underlined) 2x+5y>10, and (the ">" is underlined) x+3y<9. She need to mark off the exact area (to the nearest hundreth of a unit) with fencing. How many units of fencing will she need, to the nearest hundreth? Agent Mulder, thinking he can blanket the entire region, needs to know the area of the region,to the nearest hundreth of a unit. What is the total area of covering he will need?

Graph the lines $y>0$, $y>\frac{-2x}{5}+2$,and $y>\frac{x}{3}+3$...and find the area of the triply shaded region

3. Hello, chibix3!

Agent Scully has discovered mysterious markings in an area bounded by:
$x \,\geq \,0,\;y \,\geq \,0,\;2x+5y\:\geq\:10,\;x+3y\:\leq\:9$

She need to mark off the region with fencing.
How many units of fencing will she need, to the nearest hundreth?

Agent Mulder needs to know the area of the region.
What is the total area, to the nearest hundredth?
$x \geq 0\text{ and }y \geq 0$ puts us in Quadrant 1.

$2x + 5y\:\geq\:10$ is the region above the line: $2x + 5y \:=\:10$
The line has intercepts: . $(5,0)\text{ and }(0,2)$
Plot the intercepts, draw the line, and shade the region above the line.

$x + 3y \:\leq \:9$ is the region below the line: $x + 3y \:=\:9$
The line has intercepts: . $(9,0)\text{ and }(0,3)$
Plot the intercepts, draw the line, and shade the region below the line.

The graph looks like this . . .
Code:
          |
3*..
|::::*..
2*:::::::::*..
|  *:::::::::::*..
|     *:::::::::::::*..
|        *:::::::::::::::*..
- - + - - - - - * - - - - - - - - * -
|           5                 9
The perimater is a messy number.

The upper (longer) diagonal is the hypotenuse of a right triangle with legs 3 and 9.
. . Its length is: . $\sqrt{3^2+9^2} \:=\:\sqrt{45} \:=\:3\sqrt{5}$

The lower (shorter) diagonal is the hypotenuse of a right triangle with legs 2 and 5.
. . Its length is: . $\sqrt{2^2 + 5^2} \:=\:\sqrt{29}$

The horizontal side has length: $4$

The vertical side has length; $1$

Therefore, the perimeter is: . $3\sqrt{5} + \sqrt{29} + 4 + 1 \:\approx\:\boxed{17.09}$

The area of the large right triangle is: . $\frac{1}{2}(9)(3) \:=\:\frac{27}{2}$

The area of the small right triangle is: . $\frac{1}{2}(5)(2) \:=\:\frac{10}{2}$

Therefore, the area of the shaded region is: . $\frac{27}{2} - 5 \:=\:\frac{17}{2} \:=\:\boxed{8.50}$