Question 323610: Please help set up and solve this problem. Thank you.
A picture frame has a length l = 13 inches and the height of h= 11 inches. If the frame width is x inches, write and expression for the display area of the frame.
I know that A=LW
Found 3 solutions by mananth, JBarnum, Theo:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! A picture frame has a length l = 13 inches and the height of h= 11 inches. If the frame width is x inches, write and expression for the display area of the frame.
L=13
h=11
..
x is the width on both sides
so you have to subtract x+x to get the length and breadth of the display area
..
display area length will be 13-2x
display area height = 11-2x
..
Area of display area = (13-2x)(11-2x)
Answer by JBarnum(2146)  (Show Source):
You can put this solution on YOUR website! this isnt a finance question second this is a volume question so V=LWH but since we are looking for just the area of the frame and not the picture we need to eliminate the picture area
(13-2x)(11-2x)= area of picture
so 13x11=143
143-((13-2x)(11-2x))=Area of frame
this is an expression:
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! L = 13
H = 11
W = x
A picture frame is normally L and W only, with no H.
If you have L and W and H then you are dealing in Volume because this is three dimensional.
With L and W only, then you are dealing in Area because this is two dimnensional.
Normally a Picture Frame has L and W only.
The actual frame part of the picture might also have an additional width, like a 1 inch border around the display area.
If this is that you mean, then you have an inner dimension to the frame and an outer dimension to the frame.
The inner dimension is the display area and the outer dimension minus the inner dimension is the frame area.
Example:
Here's a picture frame:
The outer dimensions are L = 12 and W = 12 which gives you an outer area of L * W = 12 * 12 = 144 square inches.
If the frame area is 1 inch wide, then the inner dimensions would be:
Inner L = Outer L minus 2 * the width of the frame.
Inner W = Outer W minus 2 * the width of the frame.
Since the frame is 1 inch wide, then the inner dimensions becomes:
Inner L = 12 - 2*1 = 12 - 2 = 10
Inner W = 12 - 2 * 1 = 12 - 2 = 10
The inner area becomes 10 * 10 = 100 square inches.
The frame area is equal to L * W of the inner dimensions subtracted from L * W of the outer dimensions which equals 12*12 - 10*10 = 144 - 100 = 44 square inches.
It's a little tricky to calculate the frame area directly, but this method of subtracting the area of the outer dimensions minus the area of the inner dimensions works just fine.
A picture of a frame like this would look something like the following:
12
-------------------------------------
| Frame Area |
| F ----------------------- F |
| r | 10 | r |
| a | | a |
1 | m | 1 1 | m | 1
2 | e | 0 0 | e | 2
| | Display Area | |
| A | | A |
| r | | r |
| e | | e |
| a | 10 | a |
| ----------------------- |
| Frame Area |
-------------------------------------
12
In the picture above, the display area is 10 * 10 = 100 square inches, and the total area including the frame is 12 * 12 = 144 square inches.
You can see that if the inner Length is equal to 10 inches and the outer Length is equal to 12 inches, then the difference is 2 inches which means there's an additional length of 1 inch on each side of the inner length.
Assuming this is what you are asking, then I see your problem as follows:
Picture Frame Outer L = 13 inches and Outer W = 11 inches.
What you call H is really the Width of the outer dimension of the picture frame.
I am presuming these are outer dimensions because, if they were inner dimensions, your display area would simply be L * W.
The problem states that the width of the frame is x inches.
If the outer dimension of the frame is L = 13 and W = 11, then:
The inner dimensions of the frame would have to be L = 13 - 2*x and W = 11 - 2*x.
The area of the outer dimension of the frame would be L * W = 13 * 11 = 143
The area of the inner dimension of the frame (the display area) would be:
(L-2*x) * (W-2*x) which would become (13-2*x) * (11-2*x).
There are limits as to how large the width of the frame can be.
Since the width of the display area can't be less than 0 and the length of the display area can't be less than 0, then we can find the limits easily enough.
The length of the display area which is the length of the inner dimensions is 13 - 2*x.
This means that x can't be greater than 6.5.
the width of the display area which is the width of the inner dimensions is 11 - 2*x.
This means that x can't be greater than 5.5.
Since 5.5 is smaller than 6.5, this means that x can't be greater than 5.5.
The solution to your problem is that the length and width of the display area is given by the following equation:
L(display area) = (13 - 2*x).
W(display area) = (11 - 2*X).
x cannot be greater than 5.5.
The area of the display area is equal to (13 - 2*x) * (11 - 2*x) square inches.
The area of the frame including the width of the frame is equal to 13 * 11 = 143 square inches.
The area of the frame itself (the part of the frame that surrounds the display area) is equal to the area of the frame including the width of the frame minus the area of the display area of the frame.
Remember:
The area of the frame including the width of the frame is equal to the area of the outer dimensions of the frame.
The area of the display area of the frame is equal to the area of the inner dimensions of the frame.
The outer dimensions of the frame is equal to 13 * 11 = 143 square inches.
The inner dimensions of the frame is equal to (13 - 2*x) * (11 - 2*x) square inches.
To understand how this works, let's take x = 0.
The outer dimensions are equal to L = 13 and W = 11
The inner dimensions are equal to L = 13 and W = 11
Since the frame itself (the border surrounding the display area) has 0 width, the display area of the frame is the same as the total area of the frame.
Now let's take x = 5.5.
Since 2 * x = 2 * 5.5 = 11, then:
The outer dimensions of the frame are equal to L = 13 and W = 11.
The inner dimensions of the frame are equal to (13 - 11) and W = (11 - 11).
The inner dimensions of the frame become equal to L = 2 and W = 0 which is clearly not possible since we have to have some width to the display area.
Now let's take x = some value between 0 and 5.5.
Let's try x = 3.
This makes 2 * x = 6.
The outer dimensions of the frame are equal to L = 13 and W = 11.
The inner dimensions of the frame are equal to L = (13 -6) and W = (11 - 6).
The inner dimensions of the frame become equal to L = 7 and W = 5 which is possible.
I believe the answer to your question is:
The Length of the display area of the frame is (13 - 2*x).
The Width of the display area of the frame is (11 - 2*x).
x has to be between the values of 0 and 5.5.
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