To solve this problem, we can use the formula for the area of a rectangle, which is:
A = lw
where A is the area, l is the length, and w is the width. We are given that the photo is 5 inches wide by 7 inches long, so the area of the photo is:
A_photo = lw = (5 in)(7 in) = 35 in^2
We are also given that the total area of the frame and included picture is 63 in^2. Let's call the width of the frame x. Then the length of the frame must be:
l_frame = 7 in + 2x
since there are two widths of the frame that add to the length of the photo. The width of the frame must be:
w_frame = 5 in + 2x
since there are two widths of the frame that add to the width of the photo. The area of the frame and included picture is:
A_frame = l_frame * w_frame = (7 in + 2x)(5 in + 2x)
We are given that the total area is 63 in^2, so we can set up the equation:
A_frame = A_photo + 63 in^2
(7 in + 2x)(5 in + 2x) = 35 in^2 + 63 in^2
Expanding the left side of the equation, we get:
35 in^2 + 24x + 4x^2 = 98 in^2
Subtracting 98 in^2 from both sides, we get:
24x + 4x^2 = 63 in^2 - 35 in^2 = 28 in^2
Simplifying, we get:
4x^2 + 24x - 28 = 0
Dividing both sides by 4, we get:
x^2 + 6x - 7 = 0
We can factor this equation as:
(x + 7)(x - 1) = 0
This gives us two possible solutions:
x = -7 or x = 1
Since the width of the frame cannot be negative, the only possible solution is x = 1. Therefore, the width of the frame is 1/2 inch, which is less than the maximum width that Eden wants. The frame will work for Eden.
Sketch:
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