# Thread: NEED HELP FAST! Derivitives/cost application

1. ## NEED HELP FAST! Derivitives/cost application

ok so i have this problem and i think ive got it figured just need a little help

we have a box with a square base and a volume of 4000cm^3

the sides and top cost 4.5$/m^2 the cost of the bottom is 7$/m^2

I am supposed to see how much it wil cost for minimum cost to make the box so...
Let x represent the length and width (since square base) let "y" rep height
V= 4000
4000=x^2y

y=4000/x^2

Than use surface area formula (including cost function?)
SA= 7(x^2) + 4.50(x^2) + 4.50 (4xy)
=11.50x^2 +18x(4000/x^2) (subbed Y value in from volume equation)
=11.50x^2 + 72000/x

than take derivitive
23x -72000/x^2

set to zero

23x^3 = 72000
x=14.6286?...

one more problem i have in the end..now how do i solve the problem of the cm^2 and the M^2 should i just divide the 4000 by 10000( or 1000 im not sure?) immediately or how should i deal with that?

2. Originally Posted by jamman790
ok so i have this problem and i think ive got it figured just need a little help

we have a box with a square base and a volume of 4000cm^3

the sides and top cost 4.5$/m^2 the cost of the bottom is 7$/m^2

I am supposed to see how much it wil cost for minimum cost to make the box so...
Let x represent the length and width (since square base) let "y" rep height
V= 4000
4000=x^2y

y=4000/x^2

Than use surface area formula (including cost function?)
SA= 7(x^2) + 4.50(x^2) + 4.50 (4xy)
=11.50x^2 +18x(4000/x^2) (subbed Y value in from volume equation)
=11.50x^2 + 72000/x

than take derivitive
23x -72000/x^2

set to zero

23x^3 = 72000
x=14.6286?...

one more problem i have in the end..now how do i solve the problem of the cm^2 and the M^2 should i just divide the 4000 by 10000( or 1000 im not sure?) immediately or how should i deal with that?
I would convert the units to m^3 at the very beginning. 4000 cm^3 into m^3

There are 100 * 100 * 100 = 1 000 000 cm^3 in a m^3. The work looks ok, except for the units.

Good luck!!