Question

what is the area of the bounded region |x|+|4y|=16

Answer 1

I think I would start by rearrange the eqn -> |x| = 16 - |4y|. and then sketch the four bounding curves for x,-x and y,-y.

Answer 2

Can you so |4y|=-|x|+16 also?

Answer 3

There are 4 straight lines 4y+x=16,x-4y=16,x-4y+16=0, x+4y+16=0

Answer 4

There are 4 straight lines 4y+x=16,x-4y=16,x-4y+16=0, x+4y+16=0

Answer 5

Yep, both are fine.

Answer 6

amity, urs makes a rhombus, i cant do graphs

Answer 7

@Redcan What you do next

Answer 8

@freckles

Answer 9

Lets start with x = 16+4y. We know its a line so lets fine two points. When y=0 the x = ?

Answer 10

16, when x=0 y = 4

Answer 11

(16*2)*(4*2)/2?

Answer 12

thats what i think but continue on

Answer 13

yep so the first line is through (0,16), and (4,0) since if x=0 --> y=4

Answer 14

Then 1st two lines intersect yand xaxis at (0,4) (16,0) (0,-4) .(16,0) common . Now 2nd two at (0,4) (0,-4) and (-16,0).now think that two lines pairs are perpendicular....hence it is a square of side length[ (16-0)^2+(0-4)^2] area is side square that is 272square unit

Answer 15

Then 1st two lines intersect yand xaxis at (0,4) (16,0) (0,-4) .(16,0) common . Now 2nd two at (0,4) (0,-4) and (-16,0).now think that two lines pairs are perpendicular....hence it is a square of side length[ (16-0)^2+(0-4)^2] area is side square that is 272square unit

Answer 16

Yep, we were only at the draw the picture stage

Answer 17

it says the answer is 128?

Answer 18

oh i get it now, the guy witht he 4 equations was right, graph it and u ave 16*2 and 4*2 as P and Q, PQ/2 = 128

Answer 19

Yes it is not a square area=4(1/2•16•4=128 i have mistaken there

Answer 20

Yes it is not a square area=4(1/2•16•4=128 i have mistaken there

Answer 21

rhombus area = PQ/2, P = 32, Q=8, when u graph it u see side lengths, u were right