Thread: Solving College Calculus 1 Problems?

I'm having a lot of issues in trying to solve applied Calculus problems (for college calc 1 classes). My biggest issue is trying to figure out some sort of methodology for figuring out what I'm supposed to do.

Here's the example of a problem i'm stuck with right now:

A closed rectangular container with a square base is to have a volume of 2000 centimeters cubed. It costs twice as much per square centimeter for the top and bottom as it does for the sides. Find the dimensions of the container of least cost.

X (squared)
----------
|0000000|
|0000000|
|0000000| Y (Of course this is 3 dimensional rectangle, but I did my best)
|0000000|
|0000000|
|0000000|
----------

So from what I understand you need LEAST cost. But I have no idea how to set up an equation for cost. I Understand:

2Y = X
V = X(squared)Y
and I have a 2000 centimeter cubed volume

From here i'm literally stuck, what equation do I "make" to substitute for a variable into the Volume equation? And am I taking the derivative of Volume in order to solve for a particular variable? Any help on this matter would be greatly appreciated. Thanks.

surface area,

cost ...



substitute for y ...





find and minimize the cost

Hello, Tactified!

I'll give you a complete walk-through . . . with baby-steps.

A closed rectangular box with a square base is to have a volume of 2000 cm³.
It costs twice as much per cm² for the top and bottom as it does for the sides.
Find the dimensions of the container of least cost.

Code:

         *- - - -*
        /       /|
       /       / |
      * - - - *  |y
      |       |  |
      |       |  |
     y|       |  *
      |       | /x
      |       |/
      * - - - *
          x

The volume of the box is 2000 cm³: . .[1]

The four sides of the box have an area of cm² each: .
Let = price of material per cm² for the sides.
. . dollars.

The ends (top and bottom) of the box have an area of cm² each: .
The ends cost per cm².
. . dollars.

Hence, the total cost is: . .[2]


Substitute [1] into [2]: .


Therefore, we must minimize: .

Got it?

Awww, too slow again . . . *sigh*
.

Oh ok, thanks a lot guys, I understand it now. But how would I go about figuring out such a problem when I am stuck?

For profit problems do I just assign a variable (P) to the number of things it is asking for? Like surface area in this problem was what it was dealing with?