10 replies to this topic

[kittycat]

40. A rectangular floor, having dimensions 10 m by 8 m, is decorated with a triangular rug as shown. Write a quadratic function that models the uncovered area of the floor and use it to determine the maximum uncovered area.


HOW DO I EVEN APPROACH THIS....

[Keil]

Covered Area = 80 - 8x - 10(8-x) - x(10-x). I really have no clue. All I did was just find the areas of the uncovered triangle. Once you multiple things through the equation will look like a quadratic.

After taking Applied Math II, geometry and algebra became really hard

Edited by Keil, 30 October 2011 - 04:24 PM.

[kittycat]

You need to find the vertex and all you're given are the points (0,0), (?,8) and (10,?)

[Persuasion]

Psh, who would have a triangular rug

[kittycat]

Oh wait, yeah that makes sense I can do this! Just never thought of doing that lmao, thank you.

[Rainie]

Psh, who would have a triangular rug

Lol! CX
&I was going to help.. but it seems like you've got it now? o:

[kittycat]

Fixed my mistakes but still confused with the original question

I'm getting 88 as my answer...

Edited by kittycat, 30 October 2011 - 05:57 PM.

[Jakerz]

The total area would be 10x8 which is 80m²

You have three different uncovered triangles, so you find the areas of each. Base*Height/2

1) (8 by x)/2 = 4x
2) (x by (10-x))/2 = .5(10x-x^2)
3) (10 by (8-x))/2 = 40-5x

So the quadratic formula for uncovered area would be
= (4x + 5x - ((x^2)/2) + 40 - 5x)
= (4x - ((x^2)/2) + 40)
= (4x - .5x + 40)
= 3.5x + 40

You would then set that to equal 0 to find the maximum x value, and thus the maximum uncovered area of the floor

My math in terms of quadratic equations is extremely rusty, so there's a very very good chance this is wrong, but it might give you the right idea to put you on the right track.. if not just ignore it sorry I couldn't be more help, I haven't taken math in a while

Edit: Something is wrong.. not sure what and too tired to try to figure it out right now, I'll check again in the morning.. maybe my brain will start working again

[Noitidart]

Thanks Jakerz. I love topics like these.

[iomega]

You have 3 uncovered parts:
Triangle 1 Area = 0.5*8*x
Triangle 2 Area = 0.5*(10-x)*x
Triangle 3 Area = 0.5*10*(8-x)

Total Uncovered Area = Triangle 1 Area + Triangle 2 Area + Triangle 3 Area
= 0.5*8*x + 0.5*(10-x)*x + 0.5*10*(8-x)
= 40 + 4*x - 0.5*x^2

The question asks for the maximum uncovered area, so you can plot the graph or differentiate and solve for dA/dt = 0 then sub into the uncovered area equation.
Either way, the total uncovered area is at a max when x = 4

Edited by iomega, 31 October 2011 - 04:47 AM.

[kittycat]

Thanks to all, it's great that you didn't give me the final answer I like that.

Anyway the answer is 48m^2 to anyone curious. Thanks for all the help