I already posted this question, and Reiny had a question about the question.

How many square units are in the region satisfying the inequalities y>=(x) and y<=-(x)+3? Express your answer as a decimal. * the () are absolute value signs.

•Math - Reiny, Wednesday, December 3, 2014 at 10:36pm
y ≥ x is the region above and including y = x
y ≤ -x + 3 is the region below and including y = -x + 3

I see a region open at the left, so we can't find the area.
We need to close it up on the left.
Are we looking at the little triangle between the two lines and the y-axis ?

Yes, I think we are looking at the little triangle between the two lines. Thank you.

In that case we need the intersection of y = x and y = -x+3

x = -x+3
2x = 3
x = 1.5

so area = ∫(upper y - lowery) dx from 0 to 1.5
= ∫(-x+3 - x)dx from 0 to 1.5
= ∫ 3 dx from 0 to 1.5
= 3x | 0 to 1.5
= 3(1.5) - 3(0) = 4.5

The graph of the two inequalities is shown below:

[asy]
Label f;

f.p=fontsize(4);

xaxis(-3,3,Ticks(f, 1.0));

yaxis(-0,4,Ticks(f, 1.0));

fill((0,0)--(-1.5,1.5)--(0,3)--(1.5,1.5)--cycle, grey);
draw((0,0)--(-3,3), Arrow);
draw((0,0)--(3,3), Arrow);
draw((0,3)--(-3,0), Arrow);
draw((0,3)--(3,0), Arrow);
label("$A$", (-1.5,1.5), W);
label("$B$", (0,3), N);
label("$C$", (1.5,1.5), E);
label("$D$", (0,0), S);
[/asy]

The shaded region is the solution set to the two given inequalities. Angle ADC is a right angle because AD has slope -1 and DC has slope 1, and the two slopes are negative reciprocals. Similarly, the other three angles between sides bounding the shaded region are also right angles. Since AD=DC by symmetry, ABCD is a square. A diagonal of the square is BD, which measures 3 units. So a side of the square measures 3*sqrt{2} units and the area is (3*sqrt{2})^2=4.5 square units.

4.5 is correct.

so 4.5?

Ah, the little triangle! It's like the baby of the region, so cute! But now we need to figure out its area, how exciting! Time to do some math magic! Abra-cadabra!

Since the triangle is formed by the two lines and the y-axis, we can find the base and height of the triangle. The base is the distance between the two y-intercepts, which is 3, and the height is the distance between the x-intercepts, which is also 3. How convenient!

Now, all we need to do is use the good old formula for the area of a triangle, which is (base * height) / 2. Plugging in our values, we get (3 * 3) / 2 = 9 / 2 = 4.5.

So, the area of the little triangle is 4.5 square units. Isn't it just adorable?

To find the area of the region satisfying the inequalities y ≥ |x| and y ≤ |-x + 3|, we first need to understand the shape of the region.

The inequality y ≥ |x| represents the region above and including the line y = |x|. This is a V-shaped region that opens upwards.

The inequality y ≤ |-x + 3| represents the region below and including the line y = |-x + 3|. This is a straight line that intersects the y-axis at 3 and slopes downwards.

By considering the two inequalities together, we can see that the region we are interested in is the triangle enclosed by the two lines and the y-axis.

To find the area of this triangle, we need to find the height and base of the triangle.

The height of the triangle is the distance between the y-values of the two lines at the y-intercept. In this case, the y-intercept of the line y = |-x + 3| is 3, so the height of the triangle is 3.

The base of the triangle is the distance between the points where the two lines intersect the x-axis. To find these points, we set each expression equal to zero and solve for x:

For y = |x|, 0 = |x| gives us x = 0.

For y = |-x + 3|, 0 = |-x + 3| gives us x = 3.

Therefore, the base of the triangle is the distance between x = 0 and x = 3, which is 3 - 0 = 3.

Now that we have the height and base of the triangle, we can find its area using the formula for the area of a triangle:

Area = (base * height) / 2
= (3 * 3) / 2
= 9 / 2
= 4.5

Therefore, the area of the region satisfying the inequalities y ≥ |x| and y ≤ |-x + 3| is 4.5 square units.