SOLUTION: The frame around the picture in the illustration has a constant width. how wide is the frame if the area equals the area of the picture the illustration is

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: The frame around the picture in the illustration has a constant width. how wide is the frame if the area equals the area of the picture the illustration is      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 214239This question is from textbook
: The frame around the picture in the illustration has a constant width. how wide is the frame if the area equals the area of the picture
the illustration is a square inside a squre. the frame
with a width of 12 inches and a hight of 10 inches
I hope thats enough info 2 help me

This question is from textbook

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
The frame around the picture in the illustration has a constant width. how wide is the frame if the area equals the area of the picture
the illustration is a square inside a squre. the frame
with a width of 12 inches and a hight of 10 inches
.
Draw a diagram of the problem, label all the information given it'll help you see the solution.
.
Let w = width of frame
then measure of picture is:
12-2w = width of picture
10-2w = length of picture
.
Area of picture is (length times width):
(12-2w)(10-2w)
.
Area of frame:
2(10w) + 2w(12-2w)
.
Setting them equal":
(12-2w)(10-2w) = 2(10w) + 2w(12-2w)
120-24w-20w+4w^2 = 20w + 24w - 4w^2
120-44w+4w^2 = 44w - 4w^2
8w^2+120-44w = 44w
8w^2-88w+120 = 0
w^2-11w+15 = 0
Using the quadratic equation we get two solutions:
w={9.40512483795333, 1.59487516204667}
It wouldn't make sense to have a 9+ inch width frame so the answer MUST be:
w = 1.595 inches (width of frame)
.
Details of the quadratic equation to follow:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aw%5E2%2Bbw%2Bc=0 (in our case 1w%5E2%2B-11w%2B15+=+0) has the following solutons:

w%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-11%29%5E2-4%2A1%2A15=61.

Discriminant d=61 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--11%2B-sqrt%28+61+%29%29%2F2%5Ca.

w%5B1%5D+=+%28-%28-11%29%2Bsqrt%28+61+%29%29%2F2%5C1+=+9.40512483795333
w%5B2%5D+=+%28-%28-11%29-sqrt%28+61+%29%29%2F2%5C1+=+1.59487516204667

Quadratic expression 1w%5E2%2B-11w%2B15 can be factored:
1w%5E2%2B-11w%2B15+=+1%28w-9.40512483795333%29%2A%28w-1.59487516204667%29
Again, the answer is: 9.40512483795333, 1.59487516204667. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-11%2Ax%2B15+%29