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

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.

Answer 1

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

Answer 2

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.

Answer 3

so 4.5?