SOLUTION: Hi, I am working through my Application of derivatives assignment and I have found a lot of helpful information on this website, and I am now on my final question, and this is very

Question 858243: Hi, I am working through my Application of derivatives assignment and I have found a lot of helpful information on this website, and I am now on my final question, and this is very difficult for me since it is optimization and I really get confused with optimization, this is a two part question a) and b) that I am hoping anyone can provide me with help, as it would be much appreciated
A cardboard box with a square base is to have a volume of 8 litres.
(1 litre = 1000cm^3)
a) Find the dimensions that will minimize the amount of cardboard to be used. Define your domain.
b) The cardboard for the box costs 0.1 cents/cm^2 but the cardboard for the bottom is thicker, so it costs three times as much. Find the dimensions that will minimize the cost of the cardboard.

For part a) I understand to come up with a domain and an equation and then differentiate the equation to get the critical points which I will use to get the maximum and minimum values, but I am unsure how to do this at all
For part b) I understand that it is the same as part a) but this time you are involving the cost function as well, again, I am unsure of any of this
Any help would be highly appreciated
Thank You
Answer by josgarithmetic(39616) (Show Source): You can put this solution on YOUR website!
x and x for the square bottom, and y for how tall.

, as cm^3.

Amount of cardboard would correspond to surface area of this box.
There are four sides and a top and bottom. The top and bottom are both square shaped:
A for area,
.
-
A couple ways to go from there. Substitute for x or substitute for y. Try , from the volume, and use in the AREA equation.

.

Wanting to minimize A, you want , and set equal to zero, and solve for the value of x that you need; and then determine what is y.
Derivative of A,

the 4 obviously not being zero.

-
The focus is on the NUMERATOR being zero;

.
-
Now the value for y can be found.

, which should seem to make sense.

I have not done the part (b); only did part (a). You could start this using for the cost, needing some simplification and other treatment. Maybe with the part been done, you may find a way to accomplish part b?
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