Question

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?

Answer 1

ok ill help you :)

Answer 2

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

Answer 3

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

Answer 4

Help?

Answer 5

do you need an answer?

Answer 6

or set up

Answer 7

setting up and understanding, please.

Answer 8

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

Answer 9

yes

Answer 10

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

Answer 11

is that good?

Answer 12

where did the 3x^2 come from?

Answer 13

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

Answer 14

** I meant 3x^2y

Answer 15

the length is 3 time the width

Answer 16

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

Answer 17

is that good?

Answer 18

yes

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

are you good with the surface area formula I presented?

Answer 24

If so We can proceed to the cost function

Answer 25

im sorry meters

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

allright.

Answer 30

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

Answer 31

sorry I made a slight mistake in that formula

Answer 32

the last one I wrote was correct

Answer 33

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

Answer 34

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

Answer 35

okay.

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

thank you soo much

Answer 45

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

Answer 46

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?

Answer 47

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