optimization problems: painting, rectangular bin

[wind]

Hi, I am having trouble with a few optimization problems

1. An avant-grade painting is to have a rectangular central abstract theme covering 384cm^2 of canvas with an orange 6cm margin at the top and bottom and a black 4cm margin on each side. Find the dimensions of the canvas that will cover the smallest area.

A=Lw

384=(w+12)(L-8)
(384/L-8)-12=w

if i plug this back in to the original one I will get the same thing

2.A rectangular bin, 1m high, is going to be constructed in the corner of a garage to contain odds and ends. The walls of the garage will provide two walls of the bin. If the total length of the other two walls for the bin is to be 4m, what dimensions of the bin will maximize the capacity.

so we want to find the maximum volume
v=Lwh

2L=4

I don't really understand this one, whats the other equation I'm supposed to use?

Thanks

 

[tkhunny]

1. An avant-grade painting is to have a rectangular central abstract theme covering 384cm^2 of canvas with an orange 6cm margin at the top and bottom and a black 4cm margin on each side. Find the dimensions of the canvas that will cover the smallest area.

A=Lw

384=(w+12)(L-8)

You probably can't believe how often the error is in the very beginning.

You wrote a very nice "A=Lw". You did NOT state what these values represented and immediately confused yourself.

Area of Theme = A = L*w = 384 cm^2
Area of Entire Canvas = (L+12)*(w+8)

Now you can see what you are doing.

 

[wind]

thanks, I got the width equals 24 but what do I plug that into to find the lenght?
I tyred to plug it in to

384/w=L

but then I get 16...


can some one help me with the second problem?

Thanks

 

[tkhunny]

thanks, I got the width equals 24 but what do I plug that into to find the lenght?
I tyred to plug it in to

384/w=L

but then I get 16...

If you are careful with your notation, your questions begin to answer themselves. Don't be afraid to write things out.

"I got the width equals 24" -- Great!

"plug it in to 384/w=L" -- Perfect

384/W = L ==> 384/24 = L ==> L = 16

I actually wrote "L = 16". Now there is no confusion. w = 24 and L = 16

 

[wind]

ok thanks, so it does equal 16 in the back of the book it sayed 36...so I was confused?

Can anyone help me with the second problem?

 

[tkhunny]

There is no need to be confused. If you write down the definitions and answer what the question wants, you are done.

Remember these?

Area of Theme = A = L*w = 384 cm^2
Area of Entire Canvas = (L+12)*(w+8)

We solved for L and w. Does that tell us how big the canas is? Look aat the definitions. No. L and w are for the art only. That's what the definition says.

What does the problem statement demand?

"Find the dimensions of the canvas that will cover the smallest area."

Oh. Then we don't care about the size of the art. What is the size of the Entire Canvas? Look at the definitions.