In need of assistance with this problem
Matt is hanging a painting in a Museum in Paris. The painting has 36 height and 28 width (in inches.) The wooden frame around the painting must be of uniform width. (So all sides of the frames are cut from the same wood board.) The total wall space used for this painting must equal 1505 in^2... What is the width of the frame?
Set up an equation for this problem.
What I did so farlentgh times width) lw=space of painting.
Width: 2x+28
Length: 2x+36
I then multiplied them and got 4x^2+128x+1,008
I factored it and got 4(x+14)(x+18)
And that is all I did. Can anyone help me continue to find the width? Thank you in advanced.
April 11, 2013
12 Comments • Newest first
@cchpm: Yeah I had that answer.
Thanks
[quote=ImNoMerchant]I think my teacher "mis-wrote" the problem because she showed us [url=http://i45.tinypic.com/2jtw11.png]this diagram.[/url][/quote]
I don't see a problem o.o.
@ImNoMerchant
Tried not to, but had to solve this thing xD
Factoring actually works, it gives u (2x+71)(2x-7)=0
so x=-35.5 or x=3.5
the answer is 3.5 inches
[quote=sparkshooter]Ugh I hate these type of problems. They piss me off for some reason.
I'm confused by some people's replies. Uniform width means that each edge of the painting to the edge of the frame is equal. Plus, most of the time, any type of frame usually has uniform width. Your computer screen, tv, etc.
I think the adding the 2x was correct.[/quote]
I think my teacher "mis-wrote" the problem because she showed us [url=http://i45.tinypic.com/2jtw11.png]this diagram.[/url]
Ugh I hate these type of problems. They piss me off for some reason.
I'm confused by some people's replies. Uniform width means that each edge of the painting to the edge of the frame is equal. Plus, most of the time, any type of frame usually has uniform width. Your computer screen, tv, etc.
I think the adding the 2x was correct.
[quote=bowcreaters]So no multivariable calculus involved if it's 10th grade, ok. Realistically, if you only need the answer, factor out 1505, you get 5*7*43. You would know the length needs to add up to 43, width needs to add up to 35. So length would be 3.5 inches on each side, and width would be 3.5 on each side as well. So after you get the answer, it seems like the length and the width were the same, so just set 4(x+14)(x+18)=1505.
In regards to how to solve for x, use the quadratic equation or a calculator. =)[/quote]
I'm guessing my teacher made a mistake on the uniform part because she drew a diagram similar to [url=http://i45.tinypic.com/2jtw11.png]this.[/url] (The box on the inside is the painting and the outer one is the wall.
@cchpm: Sorry I didn't see your post before posting this ^.
Thank your for the help.
What you did so far is correct, however, you don't factor 4x^2+128x+1,008
What you do is make 4x^2+128x+1,008=1505
then subtract 1505
4x^2+128x-497=0
use quadratic equation or factoring(not sure if it works) to solve for x
you should get a positive value & a negative value, you know which is the answer =P
EDIT: just a little tip, in order to solve this kind of equation, factoring it 1st will complicate everything. So just move everything to 1 side & make it equal to 0. Then solve for x
So no multivariable calculus involved if it's 10th grade, ok. Realistically, if you only need the answer, factor out 1505, you get 5*7*43. You would know the length needs to add up to 43, width needs to add up to 35. So length would be 3.5 inches on each side, and width would be 3.5 on each side as well. So after you get the answer, it seems like the length and the width were the same, so just set 4(x+14)(x+18)=1505.
In regards to how to solve for x, use the quadratic equation or a calculator. =)
Well, it actually isn't a rectangular frame. It would actually be a rectangular frame, just with smoothed corners. I hope I complicated your problem!
[quote=bowcreaters]1) When you put x you are assuming that the frame width and frame length are the same. But the problem says only uniform width. So maybe it could be 2x+28, 2y+36?
2) What grade math is this?
3) If the total area was divisible by 4, I would just equate 4(x+14)(x+18) = 1505, but it is not.[/quote]
1) Oh yeah, I didn't consider that. One moment please
2) 10th
1) When you put x you are assuming that the frame width and frame length are the same. But the problem says only uniform width. So maybe it could be 2x+28, 2y+36?
2) What grade math is this?
3) If the total area was divisible by 4, I would just equate 4(x+14)(x+18) = 1505, but it is not.
[quote=danman]Did Matt paint the painting or is he hanging up someone else's artwork?[/quote]
I guess he is hanging up someone else's artwork since artists usually don't hang* their own paintings in museums. o.o