SOLUTION: can you please help me with this word problem?
you have a rectangular piece of wrapping paper whose perimeter is at most 20 inches. You want to use it to cover a cubical box of l
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you have a rectangular piece of wrapping paper whose perimeter is at most 20 inches. You want to use it to cover a cubical box of l
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Question 425102: can you please help me with this word problem?
you have a rectangular piece of wrapping paper whose perimeter is at most 20 inches. You want to use it to cover a cubical box of length of two inches. sketch the graph that shows the possible area of the paper. can you cover the box?
i really don't under stand where to start. I tried multiplying 2(inches) by 6(sides) and getting 12 but i don't get how to graph the area.
thank you in advance
izzy Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! You have a rectangular piece of wrapping paper whose perimeter is at most 20 inches. You want to use it to cover a cubical box of length of two inches. sketch the graph that shows the possible area of the paper. can you cover the box?
i really don't under stand where to start. I tried multiplying 2(inches) by 6(sides) and getting 12 but i don't get how to graph the area.
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Each of the 6 faces of the cube is 2"x2" = 4 sq.in.
6*4 sq.in. = 24 in^2 is the area of wrapping paper needed to cover the cube.
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If the perimeter of the paper is 20".
Perimeter = 2(length + width)
20 = 2(length+*width)
length + width = 10 in
length = 10-width
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Area = length*width
Area = (10-w)(w)
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Solve: w(10-w) >= 24
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-w^2+10w-24 >= 0
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w^2-10w+24<= 0
(w-6)(w-4) <= 0
Positive solution:
width = 6 inches
length = 4 inches
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Conclusion: The area of the paper is large enough
cover the cube.
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Cheers,
Stan H.
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