An open box is being constructed whose base length is 3 times the base width and whose volume is 50 cubic meters. If thematerials used to build the box cost $10 per square meter for the bottom and $6 per square meter for the sides, what are the dimensions for the least expensive box? Help?!?! I feel as if not enough information is provided. Yet again I could be wrong? How do I solve this problem?

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An open box is being constructed whose base length is 3 times the base width and whose volume is 50 cubic meters. If thematerials used to build the box cost $10 per square meter for the bottom and $6 per square meter for the sides, what are the dimensions for the least expensive box? Help?!?! I feel as if not enough information is provided. Yet again I could be wrong? How do I solve this problem?

Mathematics
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ok ill help you :)
this is an optimization problem, did you see that in class?
Yes I do recognize that its an optimization problem, in which we have to alter our formulas.

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Other answers:

Help?
do you need an answer?
or set up
setting up and understanding, please.
okay, so first lets take care of the constraint. i.e. the volume of this box is 50 cm^3
yes
how about \[V=3x^2y\] and we know that \[50=3x^2y\]
is that good?
where did the 3x^2 come from?
woah sorry i had to go fix a computer ill let ebbflo take over hes got a good flow going,
** I meant 3x^2y
the length is 3 time the width
so 3x^2 is the area of the base and I just let y be the heigth of the box
is that good?
yes
okay now we need the surface area formula for this box which since there is no top is \[S=3x(x)+2(3x)y+2xy\]
Area = base * height so the base =3x and the height=x
i am just adding up the areas of the sides of the box
the base of the box has dimensions (3x)cm by (x)cm
are you good with the surface area formula I presented?
If so We can proceed to the cost function
im sorry meters
can you explain to me how you arrived to your Surface area
since this box only has 5 sides we just add up the areas of those sides the area of the bottom is (3x)(x) the combined area of one of the pairs of equal sides is 2(3x)(y) and the combined area of the remaining pair of equal sides is 2(x)(y) then I just summed those up
so that's where I got my formula for the surface area
allright.
so now ti get the cost function we just multiply 10 by the area of the bottom and 6 times each of the combined areas of the sides like so \[C(x,y)=10(3x^2)+6(6xy)+6(3xy)\] collecting like terms we pretty it up to get \[C(x,y)=30x^2+48xy\]
sorry I made a slight mistake in that formula
the last one I wrote was correct
\[C(x,y)=10(3x^2)+6(6xy)+6(2xy)\] should have been the first one from which I got \[C(x,y)=30x^2+48xy\]
now we are ready to write the cost function above as a function of x only by way of the constraint
okay.
so since \[50=3x^2y\]\[y=\frac{50}{3x^2}\]
so \[C(x)=30x^2+48x(\frac{50}{3x^2})\] and cleaning it up we get \[C(x)=30x^2+\frac{800}{x}\]
if we are good, then now it is time to derive
\[C^\prime(x)=60x-\frac{800}{x^2}\] get a common denominator to make finding the critical points easier we get\[C^\prime(x)=\frac{60x^3-800}{x^2}\]
since this is a box with known volume we know \[x\neq 0\] so just solve \[60x^3-800=0\] for x
we get \[x=2(\frac{5}{3})^\frac{1}{3}\]
this is the only value we get so many people quit there and just go with but you can confirm this is a minimum value via the second derivative test for max/min
use that vale to get the other dimensions and you are done...
thank you soo much
glad it helps, this is easier to explain with the aid of a sketch
I was just looking over the work and you told me to use my x value to figure out my other dimensions. Do i do that by taking the second derivative?
no just find the height by plugging that x value into the constraint we solved for why and the other dimension is just 3 time that x value we got

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