Can you help me on this math problem?

Question

A rectangular sheet of metal is 10cm longer than it is wide. Squares, 5 cm on a side, are cut from the corners of the sheet, and the flaps are bent up to form an open topped box having volume 6L. Find the original dimensions of the sheet of metal

Answer 1

Let the width be x cms.
Then the length is x+10 cms.

The width of the box after cutting the squares from each corners = x - 5 - 5 = x-10 cms
The length of the box = (x+10) - 5 - 5 = x cms
The height of the box = 5 cms
Therefore, the volume of the box = height * length * width
= 5 * x * (x-10)
= 5x^2 - 50x
Now, the volume given = 6L = 6 * 1000 cm^3 = 6000 cm^3

Therefore,
5x^2 - 50x = 6000
=> x^2 - 10x = 1200
=> x(x-10)= 1200

Thus x should be 40 to satisfy this equation.
Therefore, length of the metal sheet = x+10 = 40 + 10 = 50 cms
Width of the metal sheet = x = 40 cms

Answer 2

I'll get you started.

6 liters = 6000 cu cm

the height of the box is 5 cm, so the length times the width is 1200 sq cm

One side of box is X cm
the other is (X + 10) cm

So, X (X + 10) = 1200

X^2 + 10X - 1200 = 0

X = 40 and -30

so X = 40
X + 10 = 50

That is the size of the box, not the original sheetmetal. The box with the corners will add 5 cm to each end, or 10 cm to each side.

So - the final sheet size is 50 cm by 60 cm


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Answer 3

original dimention is a x (a+10)

(a-10) x (a+10-10) x 5 = (a-10)x a x5=6000

5a(a-10)-6000=0
5a^2-50a-6000=0
a^2-10a-1200=0
(a-40)(a+30)=0 a=40 or a=-30

since a has to be greater than 0 a=40

the original dimention is 40 x 50

Answer 4

(w + 10 - 10)(w - 10)(5) = 6,000
w^2 - 10w -1,200 = 0
(w + 30)(w - 40) = 0
w = 40 cm
w + 10 = 50 cm

5*30*40 = 6,000 cc = 6L