Question 116714
Desperate times call for desperate action. Here's some help. You may need to re-post this 
problem to see if another tutor can supply some more info.
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First part. The way I interpret what you wrote is that Kayla has 1.5 square meters of material.
[Note that this is NOT the same as a 1.5 meter square which is 1.5 meters on a side for
an area of 1.5^2 = 2.25 square meters.]
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There are 10,000 square centimeters in each square meter ... think that 1 square meter could
be a metal sheet 1 meter on a side. But 1 meter is 100 cm. So a sheet of metal that is 
1 meter on a side is equal to a sheet of metal that is 100 cm on a side. So a 1 meter by 1 meter
sheet of metal is equivalent to a 100 cm by 100 cm piece of material. A 100 by 100 cm sheet 
is 10,000 square cm. So the conversion from square meters to square cm is 10,000 times sq meters 
equals sq cm. 
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And since Kayla has 1.5 square meters of metal, 10,000 times that means that the metal Kayla 
has is 15,000 square centimeters of metal. 
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Second part. I have just about convinced myself that a cube is the type of box that will 
contain the maximum volume. And a cube has 6 sides (Top, Bottom, 4 Sides around). Since
a cube has equal length sides on all edges, the area of each side is S^2 and there are,
as stated, 6 sides. So the total area of material needed to make a cube is 6S^2 and this
area must come from 15,000 sq cm of metal. Solve for S by using the equation
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6S^2 = 15000
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Divide both sides by 6 and you have:
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S^2 = 2500 
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Take the square root of both sides and you have:
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S = sqrt(2500) = 50 cm
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The box with the maximum volume that can be built from 1.5 square meters of material and
having a lid is (I think) a cube having each side 50 cm long. Not a very big box for firewood. 
(That's a cube that is only about 19.6 inches by 19.6 inches by 19.6 inches.)
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Part c.
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If Kayla is going to build a box with no lid and a square bottom, the amount of material needed
for the bottom of the box will be {{{S^2}}} where S is a side of the bottom of the box. The 
question is, how tall can the box be? The material left after making the bottom of the box
is {{{15000}}} sq cm minus {{{S^2}}} cm. Each of the four sides that need to be build will be H, the height
of the box, times S in area. And since there are four sides, the total area of material needed
for the sides is {{{4*S*H}}}. Since the amount of metal left after making the bottom of the box
is {{{15000 - S^2}}}, this is the amount of material left to build the sides which requires a total
of {{{4*S*H}}}. Set these two amounts equal and solve for H:
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{{{4*S*H = 15000 - S^2}}}
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Divide both sides by 4*S and you have:
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{{{H = (15000 - S^2)/(4*S)}}}
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And we know the volume of this box is the product of its dimensions or:
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{{{V = H * S * S}}}  or {{{H*S^2}}}
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Substitute (15000 - S^2)/(4*S) for H in the volume equation to get:
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{{{V = (15000 - S^2)/(4*S)*S^2}}}
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Cancel one of S in the numerator with the S in the denominator and the equation becomes:
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{{{V = (15000 - S^2)*S/4}}}
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Now build yourself a table by assuming values for S and calculating the corresponding 
values for the Volume. 
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If I haven't made a mistake, here's a table of some values:
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When S (the dimension of the square bottom) = ......... then Volume =
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60 cm ................................................. 171,000 cm^3
65 cm ................................................. 175,094 cm^3
69 cm ................................................. 176,623 cm^3
70 cm ................................................. 176,150 cm^3
71 cm ................................................. 176,772 cm^3
72 cm ................................................. 176,688 cm^3
75 cm ................................................. 175,781 cm^3
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It looks as if the volume maximizes around the value S = 71 cm
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And when S = 71 cm the height is given by:
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{{{H = (15000 - S^2)/(4*S)= (1500- 71^2)/(4*71) = 35.0669 cm}}}
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So the dimensions of the box that appear to maximize the volume are around
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71 cm by 71 cm by 35.0669 cm where the 71 by 71 is the bottom and the 35 is the height.
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I've done this so fast I haven't had time to check it, but maybe it will help you to understand
the problem a little better and get some credit tomorrow. Check my work.
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Hope I'm not too late.