Question

Help with parts b,c and e, picture attached.

Answer 1

Possible values of r are those such that the population density (P) > 0. If you set P = 0 and solve for r, you'll get two solutions, only one of which will make sense as the radius of the city. Anything inside that radius should be an allowable value of r.

For your graph of P vs. r, just use r as the x-axis, and P as the y-axis. You've got a quadratic with a negative coefficient on the squared term, so it's going to be an inverted parabola of some sort.

To find the point where P is greatest, find dP/dr (you already did that) and set it equal to 0. Solve that equation for r, then substitute that value of r in the original equation. The points on a curve where the first derivative is equal to 0 are inflection points, and if there is an absolute minimum or maximum, that's where you'll find it.

Answer 2

ok I understand but can you show me ur working for setting P=0 and solving it?? I did it already but I got the wrong answer.

Answer 3

Show me yours and I'll show you mine :-)

Answer 4

Did you use the quadratic formula, completing the square, or what?

Answer 5

lol k, I didn't use the quadratic formula because I couldn't.

Answer 6

P=-20r^2 + 40r + 10
0=-2r^2 + 4 + 1
-1= -2r^2 + 4r
-1 = -2r(r-2)

-2r=-1 r-2=-1
r = 1/2 or r = 1

Answer 7

shouldn't that still work... or is it wrong?

Answer 8

No, the factors have to multiply to equal 0. If you had ended up with \[-2r(r-2) = 0\] then r = 0 and r = 2 would be solutions.

Answer 9

ok, how would u simplify it?

Answer 10

You could use the quadratic formula to solve \[-2r^2+4r+1 = 0\]a = -2, b = 4, c = 1
\[r = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Or you could complete the square. Why don't you try the quadratic approach while I write up completing the square?

Answer 11

ok sure

Answer 12

To complete the square, you need to get your equation in the form \[x^2 + bx = c\]Then, take half of b, square it, and add it to both sides.\[x^2 + bx + (b/2)^2 = c+(b/2)^2\]Write the left hand side as a square:\[(x+b/2)^2 = c + (b/2)^2\]Take the square root of both sides\[(x+b/2) = \pm \sqrt{c+(b/2)^2}\]Solve for x\[x=-b/2 \pm \sqrt{c+(b/2)^2}\] If you're thinking that looks suspiciously like the quadratic formula, award yourself a gold star.

Answer 13

\[4\sqrt{2} or \frac{4\sqrt{2}+4}{4}\]

Answer 14

u are right, it does look like the quadratic formula and I know how to complete the square.

Answer 15

In the current case, we have to divide everything by -2 to get the coefficient of r^2 = 1. That gives us \[r^2 -2r -1/2 = 0\] or a = 1, b = -2, c = -1/2 in the quadratic formula.

\[r =\frac{ -(-2) \pm \sqrt{(-2)^2-4(1)(-1/2)}}{2(1)} = \frac{2\pm\sqrt{4+2}}{2} = 1\pm \frac{\sqrt{6}}{2}\]

Answer 16

u are right, but they left it at (2 +- Sqrt 6) / 2

Answer 17

I meant (2+Sqrt 6)/2

Answer 18

\[0<r \le \frac{2+\sqrt{6}}{2}\]

Answer 19

You can do the quadratic formula directly, of course, no need to factor out the 20.

Yeah, you could also write
\[1\pm\sqrt{\frac{3}{2}}\]

Answer 20

Okay, so one of those values is negative, right, you got it!

Answer 21

wait...

Answer 22

how can r be negative?

Answer 23

I'm not going anywhere...

Answer 24

and why does it have to be > than 0? it can be equal to 0 as well? because if it is r=zero then it is at it's centre.

Answer 25

Okay, the equation has two roots. One of them doesn't make sense for the problem. This is not unusual — you've got a curve (a parabola) that has been moved about so that it describes the behavior of some system, but only part of the curve is being used. Have you done the graph yet?

Answer 26

nope. I'm still confused on part b.

Answer 27

yeh but why does the answer include the negative part then?

Answer 28

Actually r can equal 0, but it must be less than 1 +- sqrt{6} because at that point P = 0.

Why don't we put the confusion about part b on hold temporarily and draw the graph. I think looking at it will help you understand what is going on. I would suggesting graphing from r = -1 through r = 4

Answer 29

okok.

Answer 30

so it will just be a negative parabola from r=-1 to r=4