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?

- anonymous

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- anonymous

ok ill help you :)

- anonymous

this is an optimization problem, did you see that in class?

- anonymous

Yes I do recognize that its an optimization problem, in which we have to alter our formulas.

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## More answers

- anonymous

Help?

- anonymous

do you need an answer?

- anonymous

or set up

- anonymous

setting up and understanding, please.

- anonymous

okay, so first lets take care of the constraint. i.e. the volume of this box is 50 cm^3

- anonymous

yes

- anonymous

how about \[V=3x^2y\] and we know that \[50=3x^2y\]

- anonymous

is that good?

- anonymous

where did the 3x^2 come from?

- anonymous

woah sorry i had to go fix a computer ill let ebbflo take over hes got a good flow going,

- anonymous

** I meant 3x^2y

- anonymous

the length is 3 time the width

- anonymous

so 3x^2 is the area of the base and I just let y be the heigth of the box

- anonymous

is that good?

- anonymous

yes

- anonymous

okay now we need the surface area formula for this box which since there is no top is \[S=3x(x)+2(3x)y+2xy\]

- anonymous

Area = base * height
so the base =3x and the height=x

- anonymous

i am just adding up the areas of the sides of the box

- anonymous

the base of the box has dimensions (3x)cm by (x)cm

- anonymous

are you good with the surface area formula I presented?

- anonymous

If so We can proceed to the cost function

- anonymous

im sorry meters

- anonymous

can you explain to me how you arrived to your Surface area

- anonymous

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

- anonymous

so that's where I got my formula for the surface area

- anonymous

allright.

- anonymous

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 \[C(x,y)=30x^2+48xy\]

- anonymous

sorry I made a slight mistake in that formula

- anonymous

the last one I wrote was correct

- anonymous

\[C(x,y)=10(3x^2)+6(6xy)+6(2xy)\] should have been the first one from which I got \[C(x,y)=30x^2+48xy\]

- anonymous

now we are ready to write the cost function above as a function of x only by way of the constraint

- anonymous

okay.

- anonymous

so since \[50=3x^2y\]\[y=\frac{50}{3x^2}\]

- anonymous

so \[C(x)=30x^2+48x(\frac{50}{3x^2})\]
and cleaning it up we get \[C(x)=30x^2+\frac{800}{x}\]

- anonymous

if we are good, then now it is time to derive

- anonymous

\[C^\prime(x)=60x-\frac{800}{x^2}\]
get a common denominator to make finding the critical points easier we get\[C^\prime(x)=\frac{60x^3-800}{x^2}\]

- anonymous

since this is a box with known volume we know \[x\neq 0\] so just solve \[60x^3-800=0\] for x

- anonymous

we get \[x=2(\frac{5}{3})^\frac{1}{3}\]

- anonymous

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

- anonymous

use that vale to get the other dimensions and you are done...

- anonymous

thank you soo much

- anonymous

glad it helps, this is easier to explain with the aid of a sketch

- anonymous

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?

- anonymous

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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